Timeline for Vector valued functions
Current License: CC BY-SA 2.5
14 events
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| Sep 17, 2010 at 3:16 | comment | added | user3103 | @Bill. If it can be disproved for coord $k$ $\leq$ degree $k$, then this would imply that the 2n-conjecture is true. The original problem is an inverse eigenvalue problem which deals with the construction of a matrix with prescribed eigenvalues (where the matrix has a specified property). The polynomial and degree constraint on $F$ comes from looking at the characteristic polynomial and viewing the coefficients as polynomials in the matrix entries. More information is available at www.uwyo.edu/bshader/notes.pdf | |
| Sep 16, 2010 at 17:38 | comment | added | Bill Thurston | @Spiked Math. Would coordinate k < degree k be good for something? If it can be disproved, would that imply the 2n-conjecture? What is your field, and what is the 2n-conjecture? It was a fun question to think about. | |
| Sep 16, 2010 at 14:48 | comment | added | user3103 | Thanks Bill for the well thought out answer! There are a few more restrictions that can be placed on $F$, such as, coordinate $k$ of $F$ has to be a polynomial of degree at most $k$. Background: The question is motivated by a known conjecture in my field, called the "2n-conjecture", which is generally believed to be true - there was an ARCC workshop on it not too long ago. My plan of attack was to relate it to a more general type of question (which I was hoping would be false). A positive answer for the question can still be useful as it gives a direction to look if seeking a counterexample. | |
| Sep 16, 2010 at 5:20 | vote | accept | CommunityBot | ||
| Sep 16, 2010 at 3:31 | history | edited | Bill Thurston | CC BY-SA 2.5 |
corrected coefficient
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| Sep 16, 2010 at 2:02 | history | edited | Bill Thurston | CC BY-SA 2.5 |
filled in gaps for all dimensions, fixed typos.; added 67 characters in body
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| Sep 15, 2010 at 23:55 | history | edited | Bill Thurston | CC BY-SA 2.5 |
Corrected degree 6 solution.
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| Sep 15, 2010 at 23:35 | history | edited | Bill Thurston | CC BY-SA 2.5 |
Solved more cases
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| Sep 15, 2010 at 23:06 | history | edited | Bill Thurston | CC BY-SA 2.5 |
added 21 characters in body; added 2 characters in body
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| Sep 15, 2010 at 22:20 | history | edited | Bill Thurston | CC BY-SA 2.5 |
added 237 characters in body
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| Sep 15, 2010 at 22:15 | comment | added | Bill Thurston | If you apply the $z \rightarrow z^4$ map to the first two coordinates, the image is an $\mathbb R^2 \times $ an orthant in the other coordinates. Now leave the first coordinate alone: we have at least an orthant in any other pair of coordinates. The most efficient way I see to use this technique would be to keep using disjoint pairs of coordinates until at most one is left over. So far it's degree 4. Combine the last coordinate with any one of the earlier ones, and you just have to square. This seems to make it degree 8 for $n$ odd, degree 4 for $n$ even. | |
| Sep 15, 2010 at 21:47 | comment | added | Dick Palais | Very nice argument ! I'll add a few words below on why it is false for $n = 1$. | |
| Sep 15, 2010 at 21:45 | comment | added | fedja | The degree is bounded by the dimension! Also, how do you show that you have full image when the pairs overlap? | |
| Sep 15, 2010 at 21:21 | history | answered | Bill Thurston | CC BY-SA 2.5 |