Timeline for Multiple roots of polynomials with coefficients $\pm 1$
Current License: CC BY-SA 4.0
14 events
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| Jun 12, 2022 at 13:25 | comment | added | მამუკა ჯიბლაძე | @AlexRavsky Thanks for the interesting suggestion! It indeed seems to come close. The series $1-x-x^2-x^3+x^4+x^5+x^6+x^7\prod_{n\geqslant0}(1-x^{2^n})$ has a local minimum at $\approx0.71975189$ with value $\approx-0.000975414$ | |
| Jun 12, 2022 at 11:51 | comment | added | Alex Ravsky | This MSE question suggests that a required pattern can be like Thue–Morse sequence: “Let's define the signed Thue–Morse sequence $t_n$ by the recurrence $t_0 = 1, \quad t_n = (-1)^n \, t_{\lfloor n/2\rfloor},$ or by the generating function $\sum_{n=0}^\infty t_n \, x^n=\prod_{n=0}^\infty\left(1-x^{2^n}\right)$”. | |
| Jun 12, 2022 at 7:59 | comment | added | მამუკა ჯიბლაძე | @AlexRavsky I now checked up to $\max(m,n)\leqslant10$, the only double zeros are at roots of unity. Of course $10$ is not much, but still... | |
| Jun 11, 2022 at 14:13 | comment | added | მამუკა ჯიბლაძე | @AlexRavsky Oh no sorry, there are such, starting from $m=7$. For example,$$1+x\frac{1-x-x^2-x^3-x^4+x^5+x^6+x^7}{1-x^8}=\frac{(1-x)(1+x)^2}{1+x^4}$$ | |
| Jun 11, 2022 at 13:18 | comment | added | მამუკა ჯიბლაძე | @AlexRavsky At least it seems that forcing periodicity cannot work. Eventually periodic Littlewood series are of the form $P_n+x^{n+1}\frac{P_m}{1-x^{m+1}}$ where $P_n$ and $P_m$ are Littlewood polynomials of degrees $n$ and $m$ respectively. Experimentally, it seems that none of these have any multiple roots. | |
| Jun 11, 2022 at 9:52 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 4.0 |
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| Jun 11, 2022 at 9:50 | comment | added | მამუკა ჯიბლაძე | @AlexRavsky Would be great, but I do not see how to formulate an appropriate algorithm. What to choose? Depending on what? | |
| Jun 11, 2022 at 8:17 | comment | added | Alex Ravsky | Maybe the pattern of signs can be regularized keeping convergences of $(m_n)$ to some $m_\infty$ and $(P_n(m_n))$ to $0$? I also note that if we require that $ P_n(1)$ is either −1, 0 or 1 for all $n\ge N$ if follows that for all these $n$ if $P_n(1)\ne 0$ then $P_{n+1}=P_n-P_n(1)x^{n+1}$, so it remains to provide the sign for $x^{n+1}$ only when $P_n(1)=0$, that is for odd $n$. | |
| Jun 11, 2022 at 4:40 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 4.0 |
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| Jun 11, 2022 at 4:33 | comment | added | მამუკა ჯიბლაძე | @TarasBanakh None that I could detect. It goes like 1 plus, 3 minuses, 4 pluses, 2 minuses, 2 pluses, 2 minuses, 1 plus, $1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 1,...$ | |
| Jun 11, 2022 at 4:29 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 4.0 |
updated calculations
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| Jun 11, 2022 at 4:28 | comment | added | Taras Banakh | Is any periodicity in the sequence of pluses and minuses, found by this algorithm? | |
| S Jun 10, 2022 at 22:34 | history | answered | მამუკა ჯიბლაძე | CC BY-SA 4.0 | |
| S Jun 10, 2022 at 22:34 | history | made wiki | Post Made Community Wiki by მამუკა ჯიბლაძე |