Timeline for Number of real roots of 0,1 polynomial
Current License: CC BY-SA 4.0
52 events
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| Mar 7 at 9:56 | history | edited | Fred Hucht | CC BY-SA 4.0 |
Added better solution for n=7.
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| Mar 1 at 19:23 | comment | added | Fred Hucht | @DenisIvanov See my answer below, which reduces $\nu(7)$ to be ${}\leq 41$. Note that the two found polynomials have zeroes at $x=-1$. | |
| Mar 1 at 19:17 | answer | added | Fred Hucht | timeline score: 3 | |
| Feb 21 at 13:37 | answer | added | fedja | timeline score: 11 | |
| Jan 14 at 4:49 | answer | added | fedja | timeline score: 16 | |
| Jan 11 at 21:20 | history | edited | Peter Mueller | CC BY-SA 4.0 |
Edited table
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| Jan 11 at 19:01 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 11 at 18:53 | comment | added | Denis Ivanov | @CommandMaster, yes, you are right, I've formulated more correctly the observation about the behavior on $[-1, 0]$ | |
| Jan 11 at 18:51 | comment | added | Denis Ivanov | @Wolfgang, thank you! I’m now a little obsessed with $0,1$-polys (o_O) so all ideas is interesting! | |
| Jan 11 at 18:47 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 11 at 18:29 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 10 at 13:46 | comment | added | Daniel Weber | The approximation doesn't seem very significant - the polynomials $x^n + x^{n+1}$ are much better approximators | |
| Jan 10 at 9:03 | comment | added | Wolfgang | Fascinating question. Long ago, I asked in a similar vein about the largest complex root of a $\pm1$-polynomial. That one shows surprisingly self-similar patterns. No idea whether it is relevant here, but who knows? | |
| Jan 9 at 22:17 | answer | added | fedja | timeline score: 20 | |
| Jan 8 at 18:08 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 8 at 18:02 | comment | added | Denis Ivanov | @YCor, yes, I post proposal just to improve A362344 with a new data. But it's not "surprisingly", 'cause after $\nu(5)$ it takes a lot of calculations =) | |
| Jan 8 at 14:25 | history | became hot network question | |||
| Jan 8 at 10:25 | comment | added | YCor | Indeed, OEIS $f(n)=A362344(n)$ (thanks @CommandMaster) is more natural to define, since $\nu$ is just derived from $f$ as $\nu(m)=\inf\{n:f(n)=m\}$, and since practically computing $\nu$ involves computing $f$. For instance if you computed $\nu(6)=28$, it probably means you computed $f(n)$ for all (even) $n\le 28$. Surprisingly OEIS currently only provides $f(n)$ for $n\le 19$ and it means it could be easily improved. | |
| Jan 8 at 6:13 | comment | added | Denis Ivanov | @GerryMyerson Thank you, fix it | |
| Jan 8 at 6:07 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 7 at 22:54 | answer | added | Peter Mueller | timeline score: 14 | |
| Jan 7 at 19:42 | comment | added | Gerry Myerson | The first paragraph makes it sound like $\nu(n)$ is the number of roots, when it's actually the degree, and $n$ is the number of roots. | |
| Jan 7 at 12:12 | comment | added | Daniel Weber | @YCor Yes, that's A362344 | |
| Jan 7 at 10:32 | comment | added | YCor | The function mapping $d$ to the maximal number of roots of such a polynomial would maybe be more natural to consider (and contains strictly more information). I you know its first values (say, about 30), you might try to locate it on OEIS. | |
| Jan 7 at 6:52 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 7 at 6:46 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 7 at 4:57 | comment | added | Timothy Chow | Another (extremely popular) MO question about Newman polynomials: Why do polynomials with coefficients 0,1 like to have only factors with 0,1 coefficients? | |
| Jan 6 at 14:50 | comment | added | Max Horn | @PeterMueller cool! Could you elaborate a bit on how you found that and how you know it is the lexicographically smallest, perhaps in an answer? | |
| Jan 6 at 14:46 | answer | added | Timothy Chow | timeline score: 31 | |
| Jan 6 at 9:14 | comment | added | Peter Mueller | The lexicographically smallest 0-1-polynomial with $6$ real roots is $x^{28} + x^{27} + x^{25} + x^{20} + x^{18} + x^{14} + x^{13} + x^{11} + x^{9} + x^{8} + x^{7} + x^{4} + x^{2} + x$, so $\nu(6)=28$. | |
| Jan 6 at 7:52 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 6 at 7:06 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 6 at 6:55 | comment | added | Denis Ivanov | @kerzol, Yes I saw it, but no! Some guys from Julia community scan up to 25 degree, post updated | |
| Jan 5 at 23:37 | comment | added | kerzol | @DenisIvanov maybe it's oeis.org/A309805 ? :) | |
| Jan 5 at 19:41 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 5 at 19:22 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 5 at 19:09 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 5 at 18:53 | comment | added | Denis Ivanov | Dear @te4, yes I think in that direction, so we need some smart filtration tools | |
| Jan 5 at 18:53 | comment | added | Robert Israel | Well, it's not A336903. A random search found $x^{38}+x^{37}+x^{35}+x^{33}+x^{32}+x^{30}+x^{29}+x^{28}+x^{26}+x^{24}+x^{17}+x^{15}+x^{11}+x^{10}+x^{9}+x^{8}+x^{7}+x^{4}+x^{2}+x$ with $6$ real roots. | |
| Jan 5 at 18:39 | comment | added | te4 | My 2¢ reducing the problem to a purely combinatorial: the number of real roots of $\sum_0^n a_kx^k=\prod(x-\alpha_k)$ (roots are pairwise distinct) is $e_+-e_-$ where $e_+$ is the number of positive eigenvalues of $(s_{i+j-2})_{i,j=1}^n$ and $e_-$ is the number of its negative eigenvalues. Here $s_m=\sum\alpha_k^m$ is a power-sum polynomial. You can compute them via Newton identities, but it seems unreasonably hard for large enough number of nonzero coefficients. Brute force is helpless here, I guess. | |
| Jan 5 at 18:20 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 5 at 18:09 | comment | added | Denis Ivanov | @CommandMaster Hmm, I try, but can't find direct pattern of behaviour =( | |
| Jan 5 at 17:54 | comment | added | Denis Ivanov | @MaxAlekseyev, yes, I think about it, so match with A336903 may be wrong. But anyway looks like there is some low under this subject | |
| Jan 5 at 17:31 | comment | added | Daniel Weber | To preserve the pattern of each polynomial containing the previous one, for 5 roots there's $p_5(x) = x^{19} + x^{18} + x^{16} + x^{13} + p_4(x)$ | |
| Jan 5 at 17:05 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 5 at 15:41 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 5 at 12:41 | comment | added | Max Alekseyev | Perhaps it's just an instance of Guy's strong law of small numbers. | |
| Jan 5 at 9:49 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag
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| Jan 5 at 9:47 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| Jan 5 at 9:47 | history | edited | Denis Ivanov | CC BY-SA 4.0 |
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| S Jan 5 at 9:46 | review | First questions | |||
| Jan 5 at 12:08 | |||||
| S Jan 5 at 9:46 | history | asked | Denis Ivanov | CC BY-SA 4.0 |