Timeline for Polynomials over $\mathbb F_2$ without zeros in $\mathbb F_2$ having an inverse series with support of large density.
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 14, 2013 at 22:27 | comment | added | Felipe Voloch | My last comment is wrong. The bound is $2^{n-1}/(2^n-1) + d/2^{2+n/2}$ | |
| Feb 14, 2013 at 21:48 | comment | added | François Brunault | @Felipe : for $n=23$ this upper bound would be approximately 0.500086, which would be $<.506$. Anyway, numerically it seems true that the larger $e$, the closer to $1/2$ are the possible densities. This is consistent with the argument using the Weil bound that you suggest in your first comment. | |
| Feb 14, 2013 at 21:33 | comment | added | Felipe Voloch | Actually, the upper bound I get is independent of $d$. Sauf erreur, it is $2^{n-1}/(2^n-1) + 1/2^{2+n/2}$. | |
| Feb 14, 2013 at 21:05 | comment | added | François Brunault | Another observation, when $n$ is prime and $2^n-1$ is composite, the density doesn't depend on $e$ alone but really depends on the chosen irreducible polynomial. | |
| Feb 14, 2013 at 21:01 | comment | added | François Brunault | Nice! I pushed the computations somewhat further and the empirical observation seems to break down : for $n=23$ and $e=178481$ the density can be $0.506\ldots$ which is $>2^{n-1}/(2^n-1)$... | |
| Feb 14, 2013 at 19:30 | comment | added | Felipe Voloch | The size of the set of $u$ in the group of order $e$ of $\mathbb{F}_{2^n}^*$ with $T(\alpha u) = 1$ is related to the number of points of the curve $y^2+y = \alpha x^d + \beta$, where $T(\beta)=1$ and $d = (2^n-1)/e$, so for large $e$ (i.e. small $d$) the Weil bound should be enough to say that about half of the $u$'s satisfy $T(\alpha u) = 1$. | |
| Feb 14, 2013 at 18:27 | history | edited | Peter Mueller | CC BY-SA 3.0 |
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| Feb 14, 2013 at 18:01 | history | answered | Peter Mueller | CC BY-SA 3.0 |