Timeline for Polynomials of low degree that clone polynomials of higher degree
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 20, 2014 at 21:16 | comment | added | Turbo | I do not really see how it works for dual forms $$f(x_1,\dots,x_{20})=x_1x_2x_3x_4x_5- x_6x_7x_8x_9x_{10}+x_{11}x_{12}x_{13}x_{14}x_{15}-x_{16}x_{17}x_{18}x_{19}x_{20}\in\Bbb R[x].$$ | |
| Dec 17, 2014 at 0:33 | vote | accept | Turbo | ||
| Dec 17, 2014 at 1:10 | |||||
| Dec 17, 2014 at 0:17 | comment | added | Turbo | Interesting.... let me read the article. | |
| Dec 16, 2014 at 22:54 | history | edited | Gjergji Zaimi | CC BY-SA 3.0 |
added 250 characters in body
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| Dec 16, 2014 at 22:30 | comment | added | Gjergji Zaimi | @Turbo it works for any product of linear forms. This is essentially the argument Alon used to prove the Komjath conjecture about covering the hypercube with hyperplanes. | |
| Dec 16, 2014 at 21:51 | comment | added | Turbo | Would the idea work if we had $(x_1+x_2-x_3+x_4)(x_5-x_6-x_7-x_8)(x_9+x_{10}-x_{11}-x_{12})(x_{13}-x_{14}+x_{15}+x_{16})$ (some terms have negative sign)? | |
| Dec 16, 2014 at 6:45 | comment | added | Turbo | Yeah you seem correct. My understanding was that $|S|\leq 3$ comes from the fact that $f=0$ if $3$ or less variables are set. It seems the same idea will work for both polynomials(for the first polynomial you can take $|S|\leq 5$ and for the second you can take $|S|\leq3$). But could you clarify the induction step further? | |
| Dec 16, 2014 at 6:37 | comment | added | Gjergji Zaimi | The proof would work the same way for all such polynomials. Try to work out an example by hand, the argument is much simpler than my notation makes it seem. | |
| Dec 16, 2014 at 6:23 | comment | added | Turbo | Also the proof may not work if we had $$f(x_1,\dots,x_{24})=(x_1+x_2+x_3+x_4)(x_5+x_6+x_7+x_8)(x_9+x_{10}+x_{11}+x_{12})(x_{13}+x_{14}+x_{15}+x_{16})(x_{17}+x_{19}+x_{19}+x_{20})(x_{21}+x_{22}+x_{23}+x_{24})\in\Bbb R[x]$$ or $$f(x_1,\dots,x_{24})=(x_1+x_2+x_3+x_4+x_5+x_6)(x_7+x_8+x_9+x_{10}+x_{11}+x_{12})(x_{13}+x_{14}+x_{15}+x_{16}+x_{17}+x_{18})(x_{19}+x_{20}+x_{21}+x_{22}+x_{23}+x_{24})\in\Bbb R[x]$$ correct? | |
| Dec 16, 2014 at 5:15 | history | answered | Gjergji Zaimi | CC BY-SA 3.0 |