Timeline for Deg $n$ integral polynomial $P(x)$ with $n+1$ integer solutions to $0\leq P\leq d$
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 6, 2019 at 21:21 | comment | added | Wolfgang | Wow, correct. Neat and clever! Now I see that @GerhardPaseman's comment about stretching has some usefulness as well. :) | |
| Jan 6, 2019 at 19:06 | comment | added | Ilya Bogdanov | @Wolfgang: Thanks, I got that luck! For $n=16$ we have the minimal $d=16!/2^{!5}=638\,512\,875$ with a polynomial $P_{16}(x)=-x^2(x^2-4)(x^2-16)(x^2-36)(x^2-64)(x^6-80x^4+1684x^2-8040)$. Double check is very welcome... | |
| Jan 6, 2019 at 15:19 | comment | added | Wolfgang | @IlyaBogdanov Wish you good luck! Well, for $n=14$, we can come very close to the lower bound, up to a factor of $\frac{1024}{1001}\approx1.023$, whereas the same for $n=16$ only yields a factor $1.074$. A bit worse than $n=10$ (factor $1.067$) - but maybe after all, the truth is closer to your lower bound than to the upper one, at least for even $n$. Funny optimisation problem! | |
| Jan 6, 2019 at 11:38 | comment | added | Ilya Bogdanov | @Wolfgang: Somehow, it seems to me that the bound may be achieved when it is integral. So I'm mostly interested now in $n=16$. I'll try to check it when I'll have a piece of paper in my hands, instead of a kid;). | |
| Jan 6, 2019 at 11:06 | comment | added | Wolfgang | @IlyaBogdanov Sure enough :) For $n=10$, best I can find is $x^2(4-x^2)(8-x^2)(16-x^2)(25-x^2)$ with $d=7560>7087.5 $ (the latter is your lower bound). | |
| Jan 6, 2019 at 8:26 | comment | added | Ilya Bogdanov | But $d=315$ for $n=8$ is exactly my lower bound! | |
| Jan 6, 2019 at 8:21 | comment | added | Wolfgang | Still better for $n=8$: $x^2(4-x^2)(8-x^2)(16-x^2)$ with $d=315<360$. | |
| Jan 6, 2019 at 7:47 | comment | added | Wolfgang | Even better: for $n=6$, take $x^2(4-x^2)(9-x^2)$ which yields $d=24$. Likewise for $n=8$, take $x^2(4-x^2)(9-x^2)(16-x^2)$, which is better than yours. The new upper bound for $n=2k$ is $d\le\prod_{j=2}^k(j^2-1)=(k-1)!(k+1)!/2$. I think generally there is still room for improvement somewhere. But I have the impression that the upper bounds of $O((\frac n2)!^2)$ are sharper than Ilya Bogdanov's lower bound. | |
| Jan 6, 2019 at 6:42 | comment | added | Haoran Chen | @Wolfgang: I checked $n=14$ and this time $(16-x^2)$ being changed to $(15-x^2)$ can improve $d$. Not sure if more and more $(4k^2-x^2)$ terms should be reduced by $1$ as even $n$ becomes larger. Not sure if it can be further reduced, either. | |
| Jan 6, 2019 at 6:19 | comment | added | Haoran Chen | @Wolfgang: I checked $n=8$ and $P_8=(1-x^2)(3-x^2)(9-x^2)(16-x^2)$ seems optimal; also checked $n=10$ and $P_8(25-x^2)$ seems optimal. The factor $(16-x^2)$ cannot be changed to $(15-x^2)$ (for $15\cdot 13\cdot 7\cdot 9>3\cdot 9\cdot 16\cdot 25$). It seems $(3-x^2)$ is the only exception. | |
| Jan 6, 2019 at 2:04 | history | edited | Haoran Chen | CC BY-SA 4.0 |
added 58 characters in body
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| Jan 6, 2019 at 2:01 | comment | added | Haoran Chen | @ChristianRemling: You are correct. I have modified the problem. | |
| Jan 5, 2019 at 21:49 | answer | added | Ilya Bogdanov | timeline score: 2 | |
| Jan 5, 2019 at 17:41 | comment | added | Gerhard Paseman | For any such polynomial P, you can scale it with a large integer K. Then KP has fewer integer solutions to the given inequalities. Gerhard "Just Stretch The Axis Some" Paseman, 2019.01.05. | |
| Jan 5, 2019 at 17:38 | comment | added | Wolfgang | In fact we can do better sometimes. For $n=6$, take $P(x)=(1-x^2)(3-x^2)(9-x^2)$ then $d=27<36$. Fascinating problem! | |
| Jan 5, 2019 at 17:23 | comment | added | Wolfgang | Note that a trivial upper bound for $n+1$ solutions is $k!^2$ for even $n=2k$ and $k!(k+1)!$ for $n=2k+1$. I wouldn't be too surprised if those bounds are sharp... | |
| Jan 5, 2019 at 15:20 | history | asked | Haoran Chen | CC BY-SA 4.0 |