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Question P. Can a polynomial $P(x)=\sum_{n=0}^ma_nx^n$ with coefficients $a_n\in\{-1,1\}$ (and $P(1)=0$) have a multiple root in the interval $(\tfrac12,1)$?

Also I am interested in a similar question for analytic functions.

Question A. Let $f(x)=\sum_{n=0}^\infty a_nx^n$ a series with coefficients $a_n\in\{-1,1\}$ such that $\sup_{m\in\mathbb N}|\sum_{n=0}^ma_i|<\infty$. Can the analytic function $f$ have a multiple root in the interval $(0,1)$?

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    $\begingroup$ If we had an example in question P without extra condition $P(1)=0$, we can reach it by multiplying $P$ by $1-x^{m+1}$. Also this yields a positive answer to question $A$ by a similar reasoning $\endgroup$
    – Fedor Petrov
    Commented Jun 10, 2022 at 12:18
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    $\begingroup$ Regarding P: even without the $P(1)=0$ restriction, up to $m=20$, there are no multiple roots at all for $m$ even, while for $m$ odd all multiple roots that may occur are $m+1$st roots of unity. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 10, 2022 at 16:02
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    $\begingroup$ Another thing about P: for each odd $m=2n-1$ there are two $P$'s with multiple roots, namely $\frac{(1-x^n)^2}{1-x}=1+x+...+x^{n-1}-x^n-x^{n+1}-...-x^{2n-1}$ and $\frac{(1-(-x)^n)^2}{1+x}=1-x+...+(-x)^{n-1}-(-x)^n-(-x)^{n+1}-...-x^{2n-1}$, and for $n$ prime these seem to be the only ones (again, checked experimentally up to $m=21$); but for $n$ not prime there are very many. Numbers of such polynomials go like $0,0,0,2,0,2,0,6,0,2,0,62,0,2,0,518,0,134,0,5452,0,2,...$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 10, 2022 at 17:48
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    $\begingroup$ Such polynomials (series) are called Littlewood polynomials (series). Vader in Theorem 7.1 of “Real roots of Littlewood polynomials” showed that a Littlewood polynomial of even degree has no real roots. $\endgroup$
    – Alex Ravsky
    Commented Jun 10, 2022 at 18:35
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    $\begingroup$ @TimothyChow, $(z^{18} + 2z^{15} + 2z^{13} + z^{12} + 2z^{11} + 3z^{10} + 3z^8 + 2z^7 + z^6 + 2z^5 + 2z^3 + z^2 + 1)(z^2 + 1)(z - 1)(z^3 - z - 1)^2$ and $(z^{18} + 2z^{15} + 2z^{13} + z^{12} + 3z^{10} + z^8 + 2z^7 + z^6 + 2z^5 + 2z^3 + z^2 + 1)(z^2 + 1)(z - 1)(z^3 - z - 1)^2$ are Littlewood polynomials with a non-cyclotomic repeated cubic factor. $\endgroup$
    – Peter Taylor
    Commented Jun 13, 2022 at 15:56

2 Answers 2

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Question P. Can a polynomial $P(x)=\sum_{n=0}^ma_nx^n$ with coefficients $a_n\in\{-1,1\}$ (and $P(1)=0$) have a multiple root in the interval $(\tfrac12,1)$?

Yes. The following four Littlewood polynomials:

  • $z^{27} + z^{26} + z^{25} + z^{24} + z^{23} - z^{22} - z^{21} + z^{20} + z^{19} + z^{18} - z^{17} - z^{16} - z^{15} - z^{14} - z^{13} - z^{12} - z^{11} - z^{10} + z^9 + z^8 - z^7 + z^6 - z^5 + z^4 - z^3 + z^2 + z - 1 = (z^{18} + z^{16} + 2z^{15} + 2z^{13} + z^{12} + 2z^{11} + 3z^{10} + 3z^8 + 2z^7 + z^6 + 2z^5 + 2z^3 + 1)(z^2 + 1)(z - 1)(z^3 + z^2 - 1)^2$
  • $z^{27} + z^{26} + z^{25} - z^{24} - z^{23} - z^{22} + z^{21} - z^{20} - z^{19} + z^{18} + z^{17} - z^{16} - z^{15} + z^{14} + z^{13} - z^{12} - z^{11} - z^{10} - z^9 - z^8 + z^7 + z^6 - z^5 + z^4 - z^3 + z^2 + z - 1 = (z^{21} - z^{20} + 2z^{19} - 2z^{18} + z^{17} + z^{16} - 3z^{15} + 3z^{14} - 2z^{13} + 2z^{11} - 4z^{10} + 4z^9 - 2z^8 - z^7 + 3z^6 - 4z^5 + 2z^4 - z^3 - z^2 + z - 1)(z^3 + z^2 - 1)^2$
  • $z^{27} + z^{26} + z^{25} + z^{24} + z^{23} - z^{22} - z^{21} + z^{20} - z^{19} - z^{18} - z^{17} + z^{16} - z^{15} + z^{14} + z^{13} - z^{12} - z^{11} - z^{10} - z^9 - z^8 + z^7 + z^6 - z^5 + z^4 - z^3 + z^2 + z - 1 = (z^{18} + z^{16} + 2z^{15} + 2z^{13} + z^{12} + 2z^{11} + z^{10} + 3z^8 + z^6 + 2z^5 + 2z^3 + 1)(z^2 + 1)(z - 1)(z^3 + z^2 - 1)^2$
  • $z^{27} + z^{26} + z^{25} - z^{24} - z^{23} - z^{22} + z^{21} + z^{20} + z^{19} + z^{18} - z^{17} + z^{16} - z^{15} - z^{14} - z^{13} + z^{12} - z^{11} - z^{10} - z^9 - z^8 + z^7 + z^6 - z^5 + z^4 - z^3 + z^2 + z - 1 = (z^{18} + z^{16} + 2z^{13} - z^{12} + 2z^{11} + z^{10} + 3z^8 + z^6 + 2z^5 + 2z^3 + 1)(z^2 + 1)(z - 1)(z^3 + z^2 - 1)^2$

all have repeated factor $z^3 + z^2 - 1$ with root $z \approx 0.75488$.

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    $\begingroup$ Did you found them by exhaustive search or some other methods? Are these of smallest degree? $\endgroup$
    – Somnium
    Commented Jun 13, 2022 at 20:43
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    $\begingroup$ @Somnium, exhaustive search, and yes. $\endgroup$
    – Peter Taylor
    Commented Jun 13, 2022 at 21:45
  • $\begingroup$ Wow. So mysterious! $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 14, 2022 at 6:11
  • $\begingroup$ I've searched as far as I'm going to go. There are 16 Littlewood polynomials of degree 27 which have repeated non-cyclotomic roots: 4 each for 4 cubic repeated factors. There are none of degrees 29, 31, or 33, which I find mildly surprising. $\endgroup$
    – Peter Taylor
    Commented Jun 28, 2022 at 7:15
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Not an answer, only some sort of evidence for A, hence community wiki. There seems to be an algorithm that produces a sequence of polynomials $(P_n)$ like in P, with $P_0=1$, $P_{n+1}=P_n\pm x^{n+1}$, $n\geqslant0$, and with $P_n$ having a local minimum at $m_n$ such that the sequence of $m_n$ converges to $\approx0.7257794$ while $P_n(m_n)$ tends to 0.

You do this: let $P_6=1-x-x^2-x^3+x^4+x^5+x^6$; it has a nice local minimum at $m_6\approx0.719842$. Now iterate the following procedure. Find minima of $P_n+x^{n+1}$, $P_n-x^{n+1}$ in the vicinity of $m_n$. Compare absolute values of these two polynomials at these minima. Choose for $P_{n+1}$ the one with smaller absolute value (and for $m_{n+1}$ where that value is attained).

Experimentally, these values become quite small quickly. For example, $P_{500}(m_{500})$ is about $1.644734\times 10^{-70}$.

Of course there is no control of the sequence $P_n(1)$ which is required to remain bounded. In fact $P_3(1)=-2$, $P_7(1)=2$, while $P_n(1)$ is either $-1$, $0$ or $1$ for all other $n$ up to $500$.

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    $\begingroup$ Is any periodicity in the sequence of pluses and minuses, found by this algorithm? $\endgroup$
    – Taras Banakh
    Commented Jun 11, 2022 at 4:28
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    $\begingroup$ @TarasBanakh None that I could detect. It goes like 1 plus, 3 minuses, 4 pluses, 2 minuses, 2 pluses, 2 minuses, 1 plus, $1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 1,...$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 11, 2022 at 4:33
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    $\begingroup$ @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. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 11, 2022 at 13:18
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    $\begingroup$ @AlexRavsky Oh no sorry, there are such, starting from $m=7$. For example,$$1+x\frac{1-x-x^2-x^3-x^4+x^5+x^6+x^7}{1-x^8}=\frac{(1-x)(1+x)^2}{1+x^4}$$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 11, 2022 at 14:13
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    $\begingroup$ @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... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 12, 2022 at 7:59

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