%%% [10] non trouvé en 1866. Article et non book %%% [19] non trouvé \documentclass[CRMATH,Unicode,biblatex,XML]{cedram} \TopicFR{Algèbre} \TopicEN{Algebra} \TopicFR{Analyse et géométrie complexes} \TopicEN{Complex analysis and geometry} \addbibresource{CRMATH_GATI_20230170.bib} \usepackage{mathtools} \newenvironment{prob}{\begin{enonce}{Problem}}{\end{enonce}} \newcommand{\lp}{\mathcal{LP}} \newcommand{\lpi}{\mathcal{LP}I} \newcommand{\pos}{\mathrm{pos}} \newcommand{\negrm}{\mathrm{neg}} \newcommand{\dd}{\mathrm{d}} \DeclareMathOperator{\sgn}{sgn} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \makeatletter \let\save@mathaccent\mathaccent \newcommand*\if@single[3]{% \setbox0\hbox{${\mathaccent"0362{#1}}^H$}% \setbox2\hbox{${\mathaccent"0362{\kern0pt#1}}^H$}% \ifdim\ht0=\ht2 #3\else #2\fi } %The bar will be moved to the right by a half of \macc@kerna, which is computed by amsmath: \newcommand*\rel@kern[1]{\kern#1\dimexpr\macc@kerna} %If there's a superscript following the bar, then no negative kern may follow the bar; %an additional {} makes sure that the superscript is high enough in this case: \newcommand*\widebar[1]{\@ifnextchar^{{\wide@bar{#1}{0}}}{\wide@bar{#1}{1}}} %Use a separate algorithm for single symbols: \newcommand*\wide@bar[2]{\if@single{#1}{\wide@bar@{#1}{#2}{1}}{\wide@bar@{#1}{#2}{2}}} \newcommand*\wide@bar@[3]{% \begingroup \def\mathaccent##1##2{% %Enable nesting of accents: \let\mathaccent\save@mathaccent %If there's more than a single symbol, use the first character instead (see below): \if#32 \let\macc@nucleus\first@char \fi %Determine the italic correction: \setbox\z@\hbox{$\macc@style{\macc@nucleus}_{}$}% \setbox\tw@\hbox{$\macc@style{\macc@nucleus}{}_{}$}% \dimen@\wd\tw@ \advance\dimen@-\wd\z@ %Now \dimen@ is the italic correction of the symbol. \divide\dimen@ 3 \@tempdima\wd\tw@ \advance\@tempdima-\scriptspace %Now \@tempdima is the width of the symbol. \divide\@tempdima 10 \advance\dimen@-\@tempdima %Now \dimen@ = (italic correction / 3) - (Breite / 10) \ifdim\dimen@>\z@ \dimen@0pt\fi %The bar will be shortened in the case \dimen@<0 ! \rel@kern{0.6}\kern-\dimen@ \if#31 \overline{\rel@kern{-0.6}\kern\dimen@\macc@nucleus\rel@kern{0.4}\kern\dimen@}% \advance\dimen@0.4\dimexpr\macc@kerna %Place the combined final kern (-\dimen@) if it is >0 or if a superscript follows: \let\final@kern#2% \ifdim\dimen@<\z@ \let\final@kern1\fi \if\final@kern1 \kern-\dimen@\fi \else \overline{\rel@kern{-0.6}\kern\dimen@#1}% \fi }% \macc@depth\@ne \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax}% \macc@set@skewchar\relax \let\mathaccentV\macc@nested@a %The following initialises \macc@kerna and calls \mathaccent: \if#31 \macc@nested@a\relax111{#1}% \else %If the argument consists of more than one symbol, and if the first token is %a letter, use that letter for the computations: \def\gobble@till@marker##1\endmarker{}% \futurelet\first@char\gobble@till@marker#1\endmarker \ifcat\noexpand\first@char A\else \def\first@char{}% \fi \macc@nested@a\relax111{\first@char}% \fi \endgroup } \makeatother \let\oldbar\bar \renewcommand*{\bar}[1]{\mathchoice{\widebar{#1}}{\widebar{#1}}{\widebar{#1}}{\oldbar{#1}}} \let\oldtilde\tilde \renewcommand*{\tilde}[1]{\mathchoice{\widetilde{#1}}{\widetilde{#1}}{\oldtilde{#1}}{\oldtilde{#1}}} %\let\tilde\widetilde %\let\hat\widehat \let\oldhat\hat \renewcommand*{\hat}[1]{\mathchoice{\widehat{#1}}{\widehat{#1}}{\oldhat{#1}}{\oldhat{#1}}} %\let\tilde\widetilde \renewcommand*{\to}{\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}} \let\oldmapsto\mapsto \renewcommand*{\mapsto}{\mathchoice{\longmapsto}{\oldmapsto}{\oldmapsto}{\oldmapsto}} \newcommand\blfootnote[1]{% \begingroup \renewcommand\thefootnote{}\footnote{#1}% \addtocounter{footnote}{-1}% \endgroup } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand*{\mk}{\mkern -1mu} \newcommand*{\Mk}{\mkern -2mu} \newcommand*{\mK}{\mkern 1mu} \newcommand*{\MK}{\mkern 2mu} \hypersetup{urlcolor=purple, linkcolor=blue, citecolor=red} \newcommand*{\romanenumi}{\renewcommand*{\theenumi}{\roman{enumi}}} \newcommand*{\Romanenumi}{\renewcommand*{\theenumi}{\Roman{enumi}}} \newcommand*{\alphenumi}{\renewcommand*{\theenumi}{\alph{enumi}}} \newcommand*{\Alphenumi}{\renewcommand*{\theenumi}{\Alph{enumi}}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \title{Further than Descartes' rule of signs} \alttitle{Au-delà de la règle des signes de Descartes } \author{\firstname{Yousra} \lastname{Gati}\IsCorresp} \address{Universit\'e de Carthage, EPT-LIM, Tunisie} \email{yousra.gati@gmail.com} %\address{Tunisia} %\email{yousra.gati@gmail.com} %\urladdr{} %% KOSTOV, Vladimir Petrov \author{\firstname{Vladimir Petrov} \lastname{Kostov}} \address{Universit\'e C\^ote d’Azur, CNRS, LJAD, France} \email{vladimir.kostov@unice.fr} %\address{France} %\email{vladimir.kostov@unice.fr} %\urladdr{} %% TARCHI, Mohamed Chaouki \author{\firstname{Mohamed Chaouki} \lastname{Tarchi}} \address[1]{Universit\'e de Carthage, EPT-LIM, Tunisie} \email{mohamedchaouki.tarchi@gmail.com} %\address{Tunisia} %\email{mohamedchaouki.tarchi@gmail.com} \keywords{\kwd{Real polynomial in one variable} \kwd{hyperbolic polynomial} \kwd{sign pattern} \kwd{Descartes' rule of signs}} \altkeywords{\kwd{Polynôme à une variable réelle} \kwd{polynôme hyperbolique} \kwd{suite de signes} \kwd{règle de Descartes}} \subjclass{26C10, 30C15} \begin{DefTralics} \newcommand*\coloneqq{:=} \newcommand\sgn{\mathrm{sgn}} \newcommand{\pos}{\mathrm{pos}} \newcommand{\negrm}{\mathrm{neg}} \end{DefTralics} %\linenumbers \begin{abstract} The \emph{sign pattern} defined by the real polynomial $Q\coloneqq \Sigma _{j=0}^da_jx^j$, $a_j\neq 0$, is the string $\sigma (Q)\coloneqq (\sgn(a_d),\ldots ,\sgn(a_0))$. The quantities $\pos $ and $\negrm $ of positive and negative roots of $Q$ satisfy Descartes' rule of signs. A couple $(\sigma _0,(\pos ,\negrm ))$, where $\sigma _0$ is a sign pattern of length $d+1$, is \emph{realizable} if there exists a polynomial $Q$ with $\pos $ positive and $\negrm $ negative simple roots, with $(d-\pos -\negrm )/2$ complex conjugate pairs and with $\sigma (Q)=\sigma_0$. We present a series of couples (sign pattern, pair $(\pos ,\negrm )$) depending on two integer parameters and with $\pos \geq 1$, $\negrm \geq 1$, which is not realizable. For $d=9$, we give the exhaustive list of realizable couples with two sign changes in the sign pattern. \end{abstract} \begin{altabstract} La suite de signes des coefficients d'un polynôme réel $Q\coloneqq \Sigma _{j=0}^da_jx^j$, $a_j\neq 0$, est donnée par $\sigma (Q)\coloneqq (\sgn(a_d),\ldots ,\sgn(a_0))$. Les quantités $\pos $ et $\negrm $ de racines positives et négatives de $Q$ satisfont la règle des signes de Descartes. Un couple $(\sigma_0,(\pos ,\negrm ))$, où $\sigma_0$ est une suite de signes de longueur $d+1$, est «~réalisable~» s'il existe un polynôme $Q$ avec $\pos $ racines simples positives et $\negrm $ racines simples négatives, avec $(d-\pos -\negrm )/2$ paires complexes conjuguées et avec $\sigma(Q)=\sigma_0$. Nous présentons une série de couples (suite de signes, paire $(\pos ,\negrm )$) dépendant de deux paramètres entiers et avec $\pos \geq 1$, $\negrm \geq 1$, qui ne sont pas réalisables. Pour $d=9$, nous donnons la liste exhaustive des couples réalisables avec deux changements de signe dans la suite de signes. \end{altabstract} \begin{document} \maketitle \section{Introduction} This paper deals with a problem which is a natural extension of Descartes' rule of signs. We consider univariate real polynomials without vanishing coefficients. About such a degree $d$ polynomial $Q$, Descartes' rule of signs states that the number $\pos $ of its positive roots is bounded by the number $c$ of sign changes in the sequence of its coefficients, the difference $c-\pos $ being even, see \cite{Ca, Cu, DG, Des, Fo, GA, J, La, Mes}. When applied to the polynomial $Q(-x)$, this rule yields the result that the number $\negrm $ of its negative roots is bounded by the number $p$ of sign preservations and the difference $p-\negrm $ is also even. In the case of \emph{hyperbolic} polynomials, i.e.~real polynomials with all roots real, one has $\pos =c$ and $\negrm =p$. Descartes' rule of signs proposes only necessary conditions. Our aim is to understand how far from sufficient they can~be. In what follows we use the quantity $\tilde{\mu}\coloneqq \min (\pos , \negrm )$ and we denote by $\tilde{\lambda}$ the number of complex conjugate pairs of roots of $Q$. Hence the quantities which we introduce satisfy the following relations in which $[.]$ denotes the integer part: \begin{equation}\label{eqdefi} \begin{array}{c@{\quad}c@{\quad}c} \pos \leq c~,&\negrm \leq p~,&\tilde{\lambda}=(d-\pos -\negrm )/2~, \\[3pt] c-\pos \in 2\mathbb{Z}~,&p-\negrm \in 2\mathbb{Z}~,& \tilde{\lambda}\leq [d/2]~, \\[3pt] \sgn (Q(0))=(-1)^{\pos }&\mathrm{and}& c+p=d~. \end{array} \end{equation} \begin{defi}\ \relax \begin{enumerate}[label=(\arabic*)] \item\label{defi1_1} A \emph{sign pattern} of length $d+1$ is a string of $d+1$ signs $+$ and/or $-$. We say that the polynomial $Q\coloneqq \sum _{j=0}^{d}a_jx^j$ defines the sign pattern $\sigma (Q)\coloneqq (\sgn (a_d)$, $\ldots$, $\sgn (a_0))$. Most often we deal with monic polynomials in which case the first sign of the sign pattern is a~$+$. \item\label{defi1_2} % (2) Suppose that a sign pattern $\sigma _0$ is given, with $c$ sign changes and $p$ sign preservations. An \emph{admissible pair} (i.e.~a pair admissible for $\sigma _0$) is a pair $(\pos , \negrm )$ satisfying conditions~\eqref{eqdefi}. In this case we say that $(\sigma _0,(\pos , \negrm ))$ is a \emph{compatible couple} (or just \emph{couple} for short). \item\label{defi1_3} % (3) We say that this compatible couple is \emph{realizable} if there exists a degree $d$ real polynomial $Q$ with $\sigma (Q)=\sigma _0$, with exactly $\pos $ positive and $\negrm $ negative roots, all of them distinct. \vspace{1mm} \item\label{defi1_4} % (4) The pair $(\pos , \negrm )=(c,p)$, which is admissible for the sign pattern $\sigma _0$, is its \emph{Descartes' pair}. \end{enumerate} \end{defi} We study the following problem: \begin{prob}\label{prob1} For a given degree $d$, which compatible couples are realizable? \end{prob} In spite of its simple formulation the problem is not trivial at all. Descartes' rule of signs gives only necessary conditions, and Problem~\ref{prob1} is a realization problem. For $d=1$, $2$ and $3$, all compatible couples are realizable, but for $d=4$, the couple $((+,-,-,-,+),(0,2))$ is not (see~\cite{Gr}). In fact, for $d=4$, this is the only non-realizable couple up to the following equivalence. \begin{defi}\label{defiZ2Z2} For a given degree $d$, the $\mathbb{Z}_2\times \mathbb{Z}_2$-\emph{action} on the set of compatible couples is defined by means of two commuting involutions. The involution \begin{equation}\label{eqim} i_m~:~(\sigma (Q), (\pos , \negrm ))\mapsto ((-1)^d\sigma (Q(-x)), (\negrm , \pos )) \end{equation} expresses the fact that the change of variable $x\mapsto -x$ changes the sign of every second coefficient and the signs of the real roots. The involution \begin{equation}\label{eqir} i_r~:~(\sigma (Q), (\pos , \negrm ))\mapsto (\sigma (Q^R(x)/Q(0))(\pos , \negrm ))~,~~~\, Q^R(x)\coloneqq x^dQ(1/x) \end{equation} is connected with the property the \emph{reverted} polynomial $Q^R$ (i.e.~$Q$ read from right to left) to have the same numbers of positive and negative roots as $Q$ (reversion changes the roots to their reciprocals). The normalizing factors $(-1)^d$ and $1/Q(0)$ are introduced to preserve the set of monic polynomials. The \emph{orbit} of each compatible couple under this action consists either of $2$ or of $4$ couples which share the same values of $\tilde{\lambda}$ and $\tilde{\mu}$ and which are simultaneously realizable or not. This is why we use the same notation for couples and for their orbits. Sometimes we consider the orbits under the $\mathbb{Z}_2\times \mathbb{Z}_2$-action only of sign patterns, not of couples. \end{defi} \begin{nota} We denote by $\Sigma _{m_1,m_2,\ldots ,m_s}$, $m_k\in \mathbb{N}$, $m_1+\cdots +m_s=d+1$, the sign pattern beginning with a sequence of $m_1$ signs $+$ followed by a sequence of $m_2$ signs $-$ followed by a sequence of $m_3$ signs $+$ etc. Example: \[ (+,+,-,-,-,+,-,+,+,+)=\Sigma _{2,3,1,1,3}~. \] \end{nota} \begin{exam} For $d=3$, the orbit of the sign pattern $\sigma ^{\diamond}\coloneqq (+,+,-,+)$ consists of four sign patterns: $\sigma ^{\diamond}$, $i_r(\sigma ^{\diamond})=(+,-,+,+)$, $i_m(\sigma ^{\diamond})=(+,-,-,-)$ and $i_r(i_m(\sigma ^{\diamond}))=(+,+,+,-)$. The orbit of the sign pattern $\Sigma _{1,d-1,1}$ ($d\geq 3$) consists of two sign patterns: \[ \Sigma _{1,d-1,1}=i_r(\Sigma _{1,d-1,1})~~~\, \mathrm{and}~~~\, i_m(\Sigma _{1,d-1,1})=i_m(i_r(\Sigma _{1,d-1,1}))=\Sigma _{2,1,\ldots ,1,2} \] ($d-3$ units). For $d$ even, both $\Sigma _{1,d-1,1}$ and $i_m(\Sigma _{1,d-1,1})$ are center-symmetric. For $d$ odd, $\Sigma _{1,d-1,1}$ is center-symmetric and $i_m(\Sigma _{1,d-1,1})$ is center-antisymmetric. \end{exam} \begin{rema}For an orbit of length $2$, one has either $i_r(\sigma (Q))=\sigma (Q)$ or $i_r(i_m(\sigma (Q)))=\sigma (Q)$. One has always $i_m(\sigma (Q))\neq \sigma (Q)$, because the second components of these two sign patterns are different. \end{rema} Problem~\ref{prob1} is formulated for the first time in \cite{AJS}. Up to the $\mathbb{Z}_2\times \mathbb{Z}_2$-action, for $d=5$ and $d=6$, there exist exactly $1$ and exactly $4$ non-realizable orbits respectively, see~\cite{AlFu}. For $d=7$ (see \cite{FoKoSh}) and $d=8$ (see \cite{FoKoSh} and \cite{KoCzMJ}) these numbers equal $6$ and $19$. In all these cases one has $\tilde{\mu}=0$ and the exhaustive answer to Problem~\ref{prob1} is known. Details are given in Section~\ref{secprelim}. For hyperbolic polynomials, the answer to Problem~\ref{prob1} is always positive, see \cite[Proposition~1]{KoSe}. In other words, the orbit of any compatible couple in which the admissible pair is the Descartes' pair is realizable. A tropical analog of Descartes' rule of signs is studied in~\cite{FoNoSh}. The first examples of non-realizable orbits with $\tilde{\mu}=1$ are found for $d=11$ in \cite{KoMB} and for $d=9$ in \cite{CGK}. For $d=9$, the orbit studied in \cite{CGK} is the only non-realizable one with $\tilde{\mu}=1$. In this paper we present a series of non-realizable orbits (depending on two integer parameters) with $\tilde{\mu}=1$. The series includes the examples for $d=11$ and $d=9$ already found. Besides, for $d=10$ and $d=11$, we give the list of the orbits with $\tilde{\mu}=1$ which are either non-realizable or for which the answer to Problem~\ref{prob1} is not known. We also show that for $d\leq 14$, with the possible exception of one orbit with $d=14$, there are no non-realizable orbits with $\tilde{\mu}\geq 2$. Finally, we give the exhaustive answer to Problem~\ref{prob1} for $d=9$, $c=2$. \begin{defi} For $d\geq 9$, we denote by $\mathcal{K}_{n,q}$ the orbit $(\Sigma _{1,n,q,1},(1,d-3))$, $n\geq 4$, $q\geq 4$, $n+q=d-1$, hence with Descartes' pair $(3,d-3)$. The orbits $\mathcal{K}_{n,q}$ and $\mathcal{K}_{q,n}$ being the same one can assume that $n\leq q$. \end{defi} \begin{theo}\label{tmmain} For $d\geq 9$, the orbit $\mathcal{K}_{n,q}$ is not realizable. \end{theo} The theorem is proved in Section~\ref{secprtmmain}. % In its proof we apply %the new results of the second author about moduli of roots of hyperbolic %polynomials, see~\cite{KoPuMaDe}. \begin{theo}\label{tmmainbis}\ \relax \begin{enumerate} \item \label{theo9_1} % (1) For $d\leq 14$, with the possible exception of the orbit $(\Sigma _{1,4,5,4,1},(2,10))$ about which the answer to Problem~\ref{prob1} is not known, there are no non-realizable orbits with $\tilde{\mu}\geq 2$. \item \label{theo9_2} % (2) For $d=10$, for the orbits $\mathcal{K}_{4,5}=(\Sigma _{1,4,5,1},(1,7))$ and $(\Sigma _{1,4,4,2},(1,7))$, one has $(\tilde{\mu},\tilde{\lambda})=(1,1)$. The first of them is not realizable (see Theorem~\ref{tmmain}), for the second one the answer to Problem~\ref{prob1} is not known. All other orbits with $d=10$ and $\tilde{\mu}=1$ are realizable. \item \label{theo9_3} % (3) For $d=11$, the following two orbits (both with $\tilde{\mu}=\tilde{\lambda}=1$) are not realizable: \[ \mathcal{K}_{5,5}=(\Sigma _{1,5,5,1},(1,8))~~~\, \mathrm{and}~~~\, \mathcal{K}_{4,6}=(\Sigma _{1,4,6,1},(1,8))~. \] For $d=11$, with the exception of these two and of the following six orbits \[ \begin{aligned} (\Sigma _{1,4,5,2},(1,8))~,~~~(\tilde{\mu},\tilde{\lambda})&=(1,1)~,& (\Sigma _{1,5,4,2},(1,8))~,~~~(\tilde{\mu},\tilde{\lambda})&=(1,1)~,\\ (\Sigma _{2,4,4,2},(1,8))~,~~~(\tilde{\mu},\tilde{\lambda})&=(1,1)~,& (\Sigma _{1,4,4,1,1,1},(1,6))~,(\tilde{\mu},\tilde{\lambda})&=(1,2)~,\\ (\Sigma _{1,4,1,1,4,1},(1,6))~,~~~(\tilde{\mu},\tilde{\lambda})&=(1,2)~,& %(\Sigma _{1,4,4,3},(1,8)~,~~~(\mu ,\lambda )=(1,1)\\ (\Sigma _{1,3,1,1,1,1,3,1},(1,4))~,~~~(\tilde{\mu},\tilde{\lambda})&=(1,3)~, \end{aligned} \] (about which the answer to Problem~\ref{prob1} is not known), all other non-realizable orbits are with $\tilde{\mu}=0$. \end{enumerate} \end{theo} The theorem is proved in Section~\ref{secprtmmainbis}. Our next result concerns orbits of sign patterns with two sign changes. Sign patterns with none or one sign change are realizable, see~\cite[Example~1 and Theorem~1]{CGK1}, and the first examples of non-realizability are obtained for $c=2$, see the beginning of Section~\ref{secprelim}. \begin{theo}\label{tm2SC} For $d=9$, the following orbits are not realizable: \[ \begin{aligned} &(\Sigma _{1,8,1},(0,k))~,~~~\, k=3~,~5~~~\, \mathrm{and}~~~\, 7~; \\ &(\Sigma _{1,7,2},(0,\ell ))~,~~~\, (\Sigma _{1,6,3},(0,\ell ))~~~\, \mathrm{and}~~~\, (\Sigma _{2,6,2},(0,\ell ))~,~~~\, \ell =5~~~\, \mathrm{and}~~~\, 7~; \\ &(\Sigma _{1,5,4},(0,7))~,~(\Sigma _{1,4,5},(0,7))~,~ (\Sigma _{2,4,4},(0,7))~,~(\Sigma _{2,5,3},(0,7))~~~\, \mathrm{and}~~~\, (\Sigma _{3,4,3},(0,7))~. \end{aligned} \] All other orbits $(\Sigma _{m,n,q},(\pos ,\negrm ))$ with $m>0$, $n>0$, $q>0$, $m+n+q=10$, are realizable. \end{theo} The theorem is proved in Section~\ref{secprtm2SC}. \section{Preliminaries\protect\label{secprelim}} We list here the non-realizable orbits for $4\leq d\leq 8$. In all cases one has $\tilde{\mu}=0$. For $d=4$ and $d=5$, the non-realizable orbits are: \begin{equation}\label{eqd4d5} % \begin{array}{lll} (\Sigma _{1,3,1},(0,2))~~~~\, \mathrm{and}~~~\, (\Sigma _{1,4,1},(0,3))~. \end{equation} For $d=6$, there are four such orbits: \begin{equation}\label{eqd=6} (\Sigma _{1,5,1},(0,2))~,~~~\, (\Sigma _{1,5,1},(0,4))~,~~~\, (\Sigma _{1,1,1,3,1},(0,2))~~~\, \mathrm{and}~~~\, (\Sigma _{2,4,1},(0,4))~. \end{equation} For $d=7$, there are six non-realizable orbits: \begin{equation}\label{eqd=7} \begin{aligned} &(\Sigma _{2,5,1},(0,5))~,&&(\Sigma _{2,4,2},(0,5))~,&& (\Sigma _{3,4,1},(0,5))~,\\ &(\Sigma _{1,4,1,1,1},(0,3))~,&& (\Sigma _{1,6,1},(0,3))~,&&(\Sigma _{1,6,1},(0,5))~. \end{aligned} \end{equation} For $d=8$, the non-realizable orbits are the following ones: \begin{equation}\label{eqd=8} \begin{aligned} &\Sigma _{2,5,2}~,~~~\, \Sigma _{1,6,2}~,~~~\, \Sigma _{1,4,4}~,~~~\, \Sigma _{1,5,3}~~~\, \mathrm{and}~~~\, \Sigma _{2,4,3}~~~\, \mathrm{with}~~~\, (0,6)~, \\ &\Sigma _{1,1,1,3,1,1,1}~~~\, \mathrm{and}~~~\, \Sigma _{1,3,1,1,1,1,1}~~~\, \mathrm{with}~~~\, (0,2)~,\\ &\Sigma _{1,5,1,1,1}~~~\, \mathrm{and}~~~\, \Sigma _{1,3,1,3,1}~~~\, \mathrm{with}~~~\, (0,2)~~~\, \mathrm{and}~~~\, (0,4)~,\\ &\Sigma _{1,7,1}~~~\, \mathrm{with}~~~\, (0,2)~,~~~\, (0,4)~~~\, \mathrm{and}~~~\, (0,6)~,\\ &\Sigma _{1,4,1,2,1}~,~~~\, \Sigma _{1,6,2}~,~~~\, \Sigma _{1,4,2,1,1}~,~~~\, \Sigma _{1,1,1,4,2} ~~~\, \mathrm{and}~~~\, \Sigma _{1,4,1,1,2}~~~\, \mathrm{with}~~~\, (0,4)~. \end{aligned} \end{equation} The following \emph{concatenation lemma} (proved in \cite{FoKoSh}) is a basic tool for proving the relizability of certain orbits. \begin{lemma}\label{lmconcat} Suppose that the monic polynomials $P_1$ and $P_2$ of degrees $d_1$ and $d_2$, with sign patterns represented in the form $(+,\sigma _1)$ and $(+,\sigma _2)$ respectively, realize the pairs $(\pos _1, \negrm _1)$ and $(\pos _2, \negrm _2)$. Here $\sigma _j$ denotes what remains of the sign patterns when the initial sign $+$ is deleted. Then \begin{enumerate} \item \label{lemma11_1} %(1) if the last position of $\sigma _1$ is $+$, then for any $\varepsilon >0$ small enough, the polynomial $\varepsilon ^{d_2}P_1(x)P_2(x/\varepsilon )$ realizes the sign pattern $(+,\sigma _1,\sigma _2)$ and the admissible pair $(\pos _1+\pos _2, \negrm _1+\negrm _2)$; \item \label{lemma11_2} %(2) if the last position of $\sigma _1$ is $-$, then for any $\varepsilon >0$ small enough, the polynomial $\varepsilon ^{d_2}P_1(x)P_2(x/\varepsilon )$ realizes the sign pattern $(+,\sigma _1,-\sigma _2)$ and the pair $(\pos _1+\pos _2, \negrm _1+\negrm _2)$. Here $-\sigma _2$ is obtained from $\sigma _2$ by changing each $+$ by $-$ and vice versa. \end{enumerate} \end{lemma} \begin{rema} We use the symbol $\ast$ to denote concatenation of couples or of sign patterns. Lemma~\ref{lmconcat} implies that when one concatenates the compatible realizable couples $(\Sigma _{m_1,\ldots ,m_s},(a,b))$ and $(\Sigma _{n_1,\ldots ,n_{\ell}},(c,d))$ one obtains the realizable couple \[ (\Sigma _{m_1,\ldots ,m_{s-1},m_s+n_1-1,n_2,\ldots ,n_{\ell}},(a+c,b+d))= ((\Sigma _{m_1,\ldots ,m_s},(a,b))\ast (\Sigma _{n_1,\ldots ,n_{\ell}},(c,d))~. \] If one considers only sign patterns instead of couples, then one can write \[ \Sigma _{m_1,\ldots ,m_{s-1},m_s+n_1-1,n_2,\ldots ,n_{\ell}}= \Sigma _{m_1,\ldots ,m_s}\ast \Sigma _{n_1,\ldots ,n_{\ell}}~. \] When necessary we use more than two consecutive concatenations. \end{rema} Consider a hyperbolic monic degree $d$ polynomial without vanishing coefficients. Suppose that the moduli of its roots are all distinct. Consider the order of these moduli on the real positive half-axis. We note only at which positions the moduli of its negative roots are; this should be clear from the following example. \begin{exam} The polynomial \[ (x-1)(x+2)(x-3)(x-4)(x+5)(x-6)(x+7)(x+8)(x-9) \] has five positive and four negative roots. We note the relative positions of the moduli of its positive and negative roots by the letters $P$ and $N$. The order of the moduli of the roots of the polynomial is \[ |1|<|-2|<|3|<|4|<|-5|<|6|<|-7|<|-8|<|9| \] which we note as \[ P0$, one can assume that one of the roots of $Q$ is at~$1$. (We remind that $Q$ has three sign changes in the sequence of its coefficients, so by Descartes' rule of signs $Q$ has either $3$ or $1$ positive roots.) We denote by $-x_i$ the negative roots of $Q$, $00$ and $a_d=1$. We explicit some of the first and some of the last coefficients~$a_j$: \begin{equation}\label{eqlist} \begin{array}{lrrrrrr} a_{d-1}&=&(e_1-1)&-&u\rlap{~,}&&\\ a_{d-2}&=&(e_2-e_1)&-&(e_1-1)u&+&v~,\\ a_{d-3}&=&(e_3-e_2)&-&(e_2-e_1)u&+&(e_1-1)v~,\\ a_{d-4}&=&(e_4-e_3)&-&(e_3-e_2)u&+&(e_2-e_1)v~,\\ a_3&=&(e_{d-3}-e_{d-4})&-&(e_{d-4}-e_{d-5})u&+&(e_{d-5}-e_{d-6})v~,\\ a_2&=&-e_{d-3}&-&(e_{d-3}-e_{d-4})u&+&(e_{d-4}-e_{d-5})v~,\\ a_1&=&&&e_{d-3}u&+&(e_{d-3}-e_{d-4})v~,\\ a_0&=&&&&&-e_{d-3}v~. \end{array} \end{equation} \let\qed\relax \end{proof} \begin{proof}[\hypertarget{(B_theo8)}{(B)}] We introduce some notation: \begin{nota} In the plane of the variables $(u,v)$ we consider the parabola $\mathcal{P}:v=u^2/4$ and the straight lines $(a_j)$ defined by the respective equations of the form $a_j=\cdots$ from the list~\eqref{eqlist}. We set \[ \begin{array}{ll} c_-\coloneqq 23-4\sqrt{30}=1.09\ldots~,&c_+\coloneqq 23+4\sqrt{30}=44.90\ldots ~, I\coloneqq (c_-,c_+)~,\\[3pt] E_1\coloneqq \sum _{\nu =1}^{d-3}1/x_{\nu}~,& E_2\coloneqq \sum _{1\leq i0$ and two negative roots if $u<0$. Polynomials below the $u$-axis have two roots of opposite signs. Our aim is to show that the domain defined by the inequalities of the form $a_j>0$ or $a_j<0$ resulting from the sign pattern $\Sigma _{1,n,q,1}$ does not intersect the domain $\{ v>u^2/4\}$. This is why in what follows we assume that $v>0$. \end{rema} We consider first the case $e_1>c_-$. As $e_1>1$ and $a_{d-1}<0$, one has $u>0$ (see~\eqref{eqlist}). For $e_1\in I$, from the definition of $e_1=x_1+\cdots +x_{d-3}$ and $E_1$ follows that \begin{equation}\label{eqeE} \begin{cases} e_1\cdot E_1\geq (d-3)^2\geq 49~~~\, \mathrm{hence}~~~\, E_1\geq 49/c_+>1~~~\, \mathrm{and}\\ e_{d-4}=e_{d-3}\cdot E_1>e_{d-3}~. \end{cases} \end{equation} \begin{lemma}\label{lmI}\ \relax \begin{enumerate}[label=(\arabic*)] \item \label{lemma16_1} %(1) For $e_1\in I$, the intersection point $S$ of the straight lines $(a_1)$ and $(a_{d-1})$ is below the parabola~$\mathcal{P}$. \item \label{lemma16_2} % Suppose that $e_1\geq c_+$ and that $x_1\leq 1$. Then the point $S$ is below the parabola~$\mathcal{P}$. \end{enumerate} \end{lemma} \begin{proof}[Proof of Lemma~\ref{lmI}]\ \relax \begin{proof}[Part~\ref{lemma16_1}] One has (see~\eqref{eqlist}) \[ (a_1)\cap (a_{d-1})~=~S~\coloneqq ~(~e_1-1~,~e_{d-3}(e_1-1)/(e_{d-4}-e_{d-3})~)~ \] and $e_{d-4}>e_{d-3}$, see~\eqref{eqeE}. The point $S$ is below the parabola $\mathcal{P}$. This follows from \[ e_{d-3}(e_1-1)/(e_{d-4}-e_{d-3})<(e_1-1)^2/4 \] which is equivalent to \begin{equation}\label{eqed34} e_{d-3}(e_1+3)e_1+3~; \] the second of these inequalities is equivalent to \begin{equation}\label{eq49} e_1^2-46e_1+49=(e_1-c_-)(e_1-c_+)<0~,~~~\, \mathrm{i.e.~to}~~~\, e_1\in I~. \end{equation} \let\qed\relax \end{proof} \begin{proof}[Part~\ref{lemma16_2}] For fixed sum $x_2+\cdots +x_{d-3}$, the sum $\sum _{\nu =2}^{d-3}1/x_{\nu}$ is minimal if $x_2=\cdots =x_{d-3}$. Hence for $x_1\leq 1$, one has \[ E_1\geq 1+(d-4)/(e_1/(d-4))=1+(d-4)^2/e_1\geq 1+36/e_1~, \] so again $e_{d-4}>e_{d-3}$. Inequality~\eqref{eqed34} can be given the equivalent form \[ 1+4/(e_1-1)1+4/(e_1-1)$ hold true for $e_1\geq c_+$ from which part~\ref{lemma16_2} of the lemma follows. \end{proof} \let\qed\relax \end{proof} There exist no couples $\mathcal{K}_{n,q}$ satisfying the assumptions of Lemma~\ref{lmI} since the lemma implies that the domain defined by the inequalities \[ %\begin{aligned} v>u^2/4, \quad a_1=e_{d-3}u+(e_{d-3}-e_{d-4})v>0\quad \mathrm{and} \quad a_{d-1}=e_1-1-u<0~,\ \ \mathrm{i.e.} \ \ u>e_1-1~, % \end{aligned} \] is void. Indeed, the straight line $(a_1)$ has a positive slope, see~\eqref{eqeE}); it intersects the parabola $\mathcal{P}$ at the origin and at a point with $u>0$. Hence the whole sector $\{ a_1>0,~u>e_1-1\}$ is below the parabola~$\mathcal{P}$. \let\qed\relax \end{proof} \begin{proof}[\hypertarget{(C_theo8)}{(C)}] Suppose that $x_{\nu}\geq 1$, $1\leq \nu \leq d-3$, so $e_1>c_-$. We consider the intersection point \[ T~\coloneqq ~(a_{d-4})\cap (a_{d-1})~=(~e_1-1~,~ ((e_3-e_2)(e_1-1)-(e_4-e_3))/(e_2-e_1)~)~. \] Observe first that for $d\geq 10$ and $x_{\nu}\geq 1$, one has $e_2-e_1>0$ and $e_3-e_2>0$. We show that the point $T$ is below the parabola $\mathcal{P}$. The straight line $(a_{d-4})$ has a positive slope. Hence the domain defined by the three inequalities \[ %\begin{array}{lll} v>u^2/4,\quad a_{d-1}=e_1-1-u<0\quad \mathrm{and}\quad a_{d-4}<0,\ \ \mathrm{i.e.}\ \ v<((e_3-e_2)u-(e_4-e_3))/(e_2-e_1),%\end{array} \] is void, so there exist no couples $\mathcal{K}_{n,q}$ with $e_1\geq c_+$ and $x_{\nu}\geq 1$. The point $T$ is under the parabola $\mathcal{P}$ exactly when \begin{align*} ((e_3-e_2)(e_1-1)-(e_4-e_3))/(e_2-e_1)<(e_1-1)^2/4~,~~~\, \intertext{i.e.} \Psi\coloneqq 4(e_3-e_2)(e_1-1)-4(e_4-e_3)-(e_1-1)^2(e_2-e_1)<0~. \end{align*} We consider the quantity $\Psi$ as a function of one of the variables $x_{\nu}$ (say, $x_1$) when the other variables $x_{\nu}$ are fixed. We denote here $\dd e_j/\dd x_1$ by $e_j'$. Clearly $e_j'=f_{j-1}$, where $f_{j-1}$ is the $(j-1)$st elementary symmetric polynomial of the quantities $x_2$, $\ldots$, $x_{d-3}$; we set $e_0=f_0=1$. Thus $e_j=f_j+x_1f_{j-1}$, \[ \begin{aligned} \Psi &=e_1^3-e_1^2e_2-2e_1^2-2e_1e_2+4e_1e_3+e_1+3e_2-4e_4\quad \mathrm{and}\\ \Psi '&=3e_1^2-2e_1e_2-e_1^2f_1-4e_1-2f_1e_1-2e_2+4f_2e_1+4e_3+1+3f_1-4f_3~. \end{aligned} \] Substituting $f_j+x_1f_{j-1}$ for $e_j$ in $\Psi '$ gives \[ \begin{aligned} \Psi '&=-(f_1+3x_1-1)(f_1^2+x_1f_1-2f_2-x_1+1)&&\\ &= -(f_1+3x_1-1)(x_1(f_1-1)+f_1^2-2f_2+1)<0~, \end{aligned} \] because $f_1>1$ and $f_2c_-$ and the point $T$ is below the parabola $\mathcal{P}$. \let\qed\relax \end{proof} \begin{proof}[\hypertarget{(D_theo8)}{(D)}] Up to now we showed that there are no couples $\mathcal{K}_{n,q}$ for $e_1>c_-$. For $e_1\in I$, this was deduced from part~\ref{lemma16_2} of Lemma~\ref{lmI}; for $e_1\geq c_+$, this follows from part~\ref{lemma16_2} of Lemma~\ref{lmI} (if $x_1\leq 1$) and from~\hyperlink{(C_theo8)}{(C)} (if $x_1>1$). Suppose now that $0c_-$; the factor $x-1$ is preserved and each of the other two factors of $Q$ is replaced by a factor of the same form. For $e_1\leq c_-$, at least $d-4$ of the quantities $x_{\nu}$ are $<1$, so after applying the involution $i_r$ they become $1/x_{\nu}>1$ and $e_1$ becomes larger than $d-4\geq 6$. Thus the proof of Theorem~\ref{tmmain} for $0c_-$.\qedhere \end{proof} \let\qed\relax \end{proof} \section{Proof of Theorem~\texorpdfstring{\protect\ref{tmmainbis}}{9}}\label{secprtmmainbis} \begin{proof}[\hypertarget{(A_theo9)}{(A)}] The proof of Theorem~\ref{tmmainbis} is organised as follows. The proof of part~\eqref{theo9_1} is given for each degree from $10$ to $14$ in \hyperlink{(B_theo9)}{(B)} -- \hyperlink{(AF_theo9)}{(F)} respectively. Parts~\hyperlink{(B_theo9)}{(B)} and~\hyperlink{(C_theo9)}{(C)} of the proof are subdivided into~\hyperlink{(B1_theo9)}{(B1)}, \hyperlink{(C1_theo9)}{(C1)} (in which we prove part~\eqref{theo9_1} of the theorem for $d=10$ and $d=11$) and~\hyperlink{(B2_theo9)}{(B2)}, \hyperlink{(C2_theo9)}{(C2)} containing the proofs of parts~\eqref{theo9_2} and~\eqref{theo9_3} of the theorem. Using the $\mathbb{Z}_2\times \mathbb{Z}_2$-action (see Definition~\ref{defiZ2Z2}) one can assume that $\pos \leq \negrm $. We remind that: \begin{enumerate}[label=(\roman*)] \item For $d\leq 8$, there are no non-realizable orbits with $\tilde{\mu}\geq 1$ (see \cite{AlFu, FoKoSh} and~\cite{KoCzMJ}). \item For $d=9$, the only non-realizable orbit with $\tilde{\mu}\geq 1$ is (see~\cite{CGK}) \begin{equation}\label{eqcased9} (\Sigma _{1,4,4,1},(1,6))~,~~~\mathrm{with}~~~\, \tilde{\mu}=1~~~\, \mathrm{and}~~~\, \tilde{\lambda}=1~. \end{equation} \item If the admissible pair of a given orbit is $(0,0)$, $(0,1)$, $(1,0)$ or $(1,1)$, then the orbit is realizable. Indeed, if the admissible pair equals $(0,0)$ or $(0,1)$ (resp. $(1,0)$), then one chooses a polynomial with the given sign pattern $\sigma$ and adds to it a sufficiently large positive (resp. negative) constant. If the admissible pair is $(1,1)$, then one represents the sign pattern in the form $\sigma =(\sigma _{\dagger},\alpha ,\beta )$, where $\alpha$ and $\beta$ are its last two signs of $\sigma$. If $\alpha =\beta$ (resp. $\alpha =-\beta$), then one uses the concatenation $(\sigma _{\dagger},\alpha ),(1,0))\ast (\Sigma _{2},(0,1))$ (resp. $(\sigma _{\dagger},\alpha ),(0,1))\ast (\Sigma _{1,1},(1,0))$). \end{enumerate} \let\qed\relax \end{proof} \begin{proof}[\hypertarget{(B_theo9)}{(B)}] Suppose that $d=10$. \begin{proof}[\hypertarget{(B1_theo9)}{(B1)}] Consider any compatible couple $\mathcal{K}_{\flat}\coloneqq (\sigma _{\flat},(\pos , \negrm ))$ with $2\leq \pos \leq \negrm $. Represent the sign pattern $\sigma _{\flat}$ as above in the form $(\sigma _{\dagger},\alpha ,\beta )$. %, where $\sigma _{\dagger}$ is the sign %pattern obtained from $\sigma _{\flat}$ by deleting its last two signs, %and $\alpha$ and $\beta$ are these signs. If $\alpha \neq \beta$ and the couple $\mathcal{K}_{\triangle}\coloneqq ((\sigma _{\dagger},\alpha ), (\pos -1, \negrm ))$ is realizable, then the couple $\mathcal{K}_{\flat}$ is also realizable as $\mathcal{K}_{\triangle}\ast (\Sigma _{1,1},(1,0))$. If $\alpha =\beta$ and the couple $\mathcal{K}_{\diamond}\coloneqq ((\sigma _{\dagger},\alpha ), (\pos , \negrm -1))$ is realizable, then one can realize $\mathcal{K}_{\flat}$ as $\mathcal{K}_{\diamond}\ast (\Sigma _{2},(0,1))$. However the couple $\mathcal{K}_{\diamond}$ is realizable. Indeed, it is with $d=9$ and either with $\tilde{\mu}=1$ when $\pos =\negrm =2$ (and there exist no such non-realizable couples) or with $\tilde{\mu}\geq 2$ when $\negrm \geq 3$, hence again realizable. So the only situation in which one does not know whether the couple $\mathcal{K}_{\flat}$ is realizable or not is when \[ \alpha \neq \beta ~~~\, \mathrm{and}~~~\, \mathcal{K}_{\triangle}=\mathcal{K}_{4,4}=(\Sigma _{1,4,4,1},(1,6))~. \] %\coloneqq ((\sigma _{\dagger},\alpha ), (\pos -1, \negrm ))= This means that $\mathcal{K}_{\flat}=(\Sigma _{1,4,4,1,1},(2,6))$ which is realizable as $(\Sigma _{1,4,4},(2,6))\ast (\Sigma _{1,1,1},(0,0))$. Hence all couples with $d=10$ and $\tilde{\mu}\geq 2$ are realizable. \let\qed\relax \end{proof} \goodbreak \begin{proof}[\hypertarget{(B2_theo9)}{(B2)}] %B2) We need some more notation: \begin{nota}\label{notaK} We denote by $\mathcal{K}_{\bullet}$ a couple with $\tilde{\mu}=1$ and by $\sigma _{\bullet}\coloneqq (\alpha _d,\ldots ,\alpha _0)$ its sign pattern, where $\alpha _j=+$ or $-$, $\alpha _d=+$. We set $\sigma ^s\coloneqq \alpha _s(\alpha _s,\ldots ,\alpha _0)$ and $\sigma _s\coloneqq (\alpha _d,\ldots ,\alpha _s)$. We discuss the possibility to realize $\mathcal{K}_{\bullet}$ as $\mathcal{K}_s\ast \mathcal{K}^s$, where the couples $\mathcal{K}_s$, $\mathcal{K}^s$ correspond to degree $d-s$ and $s$ polynomials with sign patterns $\sigma _s$ and $\sigma ^s$. \end{nota} \begin{rema}\label{remK} If $\alpha _s=+$, then the admissible pair of $\mathcal{K}_s$ (resp. of $\mathcal{K}^s$) is of the form $(0,.)$ (resp. $(1,.)$). If $\alpha _s=-$, then these admissible pairs are of the form $(1,.)$ and $(0,.)$ respectively. We remind that for $d\leq 8$, the admissible pairs of the form $(1,.)$ are realizable, see the beginning of Section~\ref{secprelim} and part~\hyperlink{(A_theo9)}{(A)} of this proof. \end{rema} Suppose first that $s=5$ and $\alpha _5=+$. Then the couple $\mathcal{K}^5$ is realizable. If $\mathcal{K}_5$ is realizable, then such is $\mathcal{K}_{\bullet}$ as well. The only possibility $\mathcal{K}_5$ not to be realizable is when $\mathcal{K}_5=(\Sigma _{1,4,1},(0,3))$, see~\eqref{eqd4d5}. So we assume that $\sigma _{\bullet}=(+,-,-,-,-,+,?,?,?,?,-)$. If $\alpha _4=-$, then the couple $\mathcal{K}_4$ is realizable and $\mathcal{K}^4$ is not realizable only when $\mathcal{K}^4=(\Sigma _{1,3,1},(0,2))$. This means that $\sigma _{\bullet}=\Sigma _{1,4,1,1,3,1}$, with Descartes' pair $(5,5)$. This sign pattern is realizable with the admissible pairs $(1,a)$, $a=3$ or~$5$: \begin{equation}\label{eqrealiz} (\Sigma _{1,4,1,1,3,1},(1,a))=((\Sigma _{1,3},(1,a-3))~\ast ~ (\Sigma _{2,1,1,3,1},(0,3))~. \end{equation} If $\alpha _5=\alpha _4=+$, then $\mathcal{K}^4$ is realizable whereas the couple $\mathcal{K}_4$ is not realizable only when it corresponds to one of the four cases listed in~(\ref{eqd=6}). In each of them one has to take into account the involution $i_r$ (see~\eqref{eqir}), but not the involution $i_m$, because the latter makes the first component of the admissible pair larger than~$1$. Hence $\sigma _4=\Sigma _{1,4,2}$. Then we consider the couples $\mathcal{K}_3$ and $\mathcal{K}^3$. The latter is always realizable, so we assume that this is not the case of $\mathcal{K}_3$. Hence $\sigma _3$ is obtained from $\sigma _4$ by adding a sign $+$ to the right, i.e.~$\sigma _3=\Sigma _{1,4,3}$. If $\sigma _2=\Sigma _{1,4,3,1}$, then both $\mathcal{K}_2$ and $\mathcal{K}^2$ are realizable, so we need to consider only the possibility $\sigma _2=\Sigma _{1,4,4}$. Thus in the end $\sigma _{\bullet}=\Sigma _{1,4,5,1}$ or $\Sigma _{1,4,4,2}$. These two sign patterns are realizable with the admissible pairs $(1,a)$, $a=3$ and~$5$: \[ \begin{aligned} (\Sigma _{1,4,5,1},(1,a))&=(\Sigma _{1,4,1},(0,1))~\ast ~ (\Sigma _{5,1},(1,a-1))~,\\ (\Sigma _{1,4,4,2},(1,a))&=(\Sigma _{1,4,1},(0,1))~\ast ~ (\Sigma _{4,2},(1,a-1))~.\end{aligned} \] The first of them is not realizable with the admissible pair $(1,7)$ (see Theorem~\ref{tmmain}), for the second one the answer is not known. Suppose now that $\alpha _5=-$. Then $\mathcal{K}_5$ is realizable while $\mathcal{K}^5$ might not be only if $\sigma ^5=\Sigma _{1,4,1}$. In this case we consider $\mathcal{K}_6$ and $\mathcal{K}^6$. For $\alpha _6=+$, the couple $\mathcal{K}^6$ is realizable whereas $\mathcal{K}_6$ could be non-realizable, if $\sigma _6=\Sigma _{1,3,1}$ hence $\sigma _{\bullet}=\Sigma _{1,3,1,1,4,1}$. This is the case~\eqref{eqrealiz} to which one has applied the involution $i_r$, so it is realizable. For $\alpha _5=\alpha _6=-$, one is, up to the involution $i_r$, in the case $\alpha _{10}=\alpha _5=\alpha _4=+$, $\alpha _0=-$, which was already considered. \let\qed\relax \end{proof} \let\qed\relax \end{proof} \begin{proof}[\hypertarget{(C_theo9)}{(C)}] Suppose that $d=11$. \begin{proof}[\hypertarget{(C1_theo9)}{(C1)}] % C1) We use the same notation as the one used in part~\hyperlink{(B1_theo9)}{(B1)} of this proof; in particular, we denote by $\mathcal{K}_{\flat}\coloneqq (\sigma _{\flat},(\pos , \negrm ))$ a compatible couple with $\tilde{\mu}\geq 2$, where $\sigma _{\flat}=(\sigma _{\dagger},\alpha ,\beta )$. As in part~\hyperlink{(B1_theo9)}{(B1)} of this proof we show that the only cases in which the couple $\mathcal{K}_{\flat}$ is possibly non-realizable are when $\alpha \neq \beta$ and \[ \mathcal{K}_{\triangle}=((\sigma _{\dagger},\alpha ),(\pos -1,\negrm ))= (\Sigma _{1,4,5,1},(1,7))=\mathcal{K}_{4,5}~~~\, \mathrm{or}~~~\, (\Sigma _{1,4,4,2},(1,7))~, \] that is, $\mathcal{K}_{\triangle}$ corresponds to one of the two cases mentioned in part~\eqref{theo9_2} of the present theorem. In these cases $\mathcal{K}_{\flat}$ is realizable, because it equals respectively \[ \begin{aligned} (\Sigma _{1,4,5,1,1},(2,7))&=(\Sigma _{1,4,1},(2,3)) \ast (\Sigma _{5,1,1},(0,4))& \mathrm{and}\\ (\Sigma _{1,4,4,2,1},(2,7))&= (\Sigma _{1,4,1},(2,3))\ast (\Sigma _{4,2,1},(0,4))~.& \end{aligned}\] \let\qed\relax \end{proof} \begin{proof}[\hypertarget{(C2_theo9)}{(C2)}] % C2) We use the notation and method of proof as developed in part~\hyperlink{(B2_theo9)}{(B2)} of this proof. We are looking for non-realizable couples $\mathcal{K}_{\bullet}$. Suppose first that % \vspace{1mm} \begin{proof}[\mdseries\hypertarget{(C2_1_theo9)}{1}] $\alpha _5=+$. Then the couple $\mathcal{K}^5$ is realizable and the couple $\mathcal{K}_5$ might not be realizable only if $\sigma _5$ corresponds to one of the cases~\eqref{eqd=6} up to the involution $i_r$. This means that $\sigma _5$ is among the sign patterns \[ \Sigma _{1,5,1}~,~~~\, \Sigma _{1,1,1,3,1}~,~~~\, \Sigma _{1,3,1,1,1}~,~~~\, \Sigma _{2,4,1}~~~\, \mathrm{and}~~~\, \Sigma _{1,4,2}~. \] \hypertarget{(C2_1.1_theo9)}{1.1.} If $\alpha _4=+$, then $\mathcal{K}^4$ is realizable and $\sigma _4$ is one of the sign patterns \[ \Sigma _{1,5,2}~,~~~\, \Sigma _{1,1,1,3,2}~,~~~\, \Sigma _{1,3,1,1,2}~,~~~\, \Sigma _{2,4,2}~~~\, \mathrm{and}~~~\, \Sigma _{1,4,3}~. \] \hypertarget{(C2_1.1.1_theo9)}{1.1.1.} The second and third sign patterns of this list do not correspond to non-realizable cases, see~\eqref{eqd=7}. \vspace{1mm} \noindent\hypertarget{(C2_1.1.2_theo9)}{1.1.2.} In the cases $\sigma _4=\Sigma _{1,5,2}$ and $\sigma _4=\Sigma _{2,4,2}$ the first component of the sign pattern $\sigma ^8$ must be~$3$ and the second must be $\geq 2$; the sign pattern $\sigma _8$ corresponds to realizable couples. Then one concludes from~\eqref{eqd=8} that $\sigma _{\bullet}$ is one of the sign patterns \[ \Sigma _{1,5,5,1}~,~~~\, \Sigma _{1,5,4,2}~,~~~\, \Sigma _{2,4,5,1}~~~\, \mathrm{and}~~~\, \Sigma _{2,4,4,2}~. \] The first of them is not realizable with the admissible pair $(1,8)$, see Theorem~\ref{tmmain}; for the second, third and fourth one the answer to this question is not known. (The second and third are in one and the same orbit.) We show the realizability of the four sign patterns with the other admissible pairs $(1,a)$, $a=6$, $4$ or~$2$: \[ \begin{aligned} (\Sigma _{1,5,5,1},(1,a))&=(\Sigma _{1,5,3},(0,a-2))~\ast ~ (\Sigma _{3,1},(1,2))\\ (\Sigma _{1,5,4,2},(1,a))&=(\Sigma _{1,5,3},(0,a-2))~\ast ~ (\Sigma _{2,2},(1,2))\\ (\Sigma _{2,4,4,2},(1,a))&=(\Sigma _{2,4,3},(0,a-2))~\ast ~ (\Sigma _{2,2},(1,2))~. \end{aligned} \] \hypertarget{(C2_1.1.3_theo9)}{1.1.3.} In the case $\sigma _4=\Sigma _{1,4,3}$ the first component of $\sigma ^8$ equals $2$ and the second is $\geq 3$. Using the list~\eqref{eqd=8} one obtains the following possibilities for~$\sigma _{\bullet}$: \[ \Sigma _{1,4,5,2}~,~~~\, \Sigma _{1,4,6,1}~,~~~\, \Sigma _{1,4,4,3}~~~\, \mathrm{and}~~~\, \Sigma _{1,4,4,1,1,1}~. \] The second of them is not realizable with the admissible pair $(1,8)$ (see Theorem~\ref{tmmain}). The third sign pattern is realizable with the admissible pair $(1,8)$. Indeed, by \cite[Theorem~3, part (iii)]{CGK1}, the couple $(\Sigma _{4,4,3},(0,8))$ is realizable, so one can set \[ (\Sigma _{1,4,4,3},(1,8))=((+,-),(1,0))\ast (\Sigma _{4,4,3},(0,8))~. \] The realizability of other possible cases is given below; for the cases which are not covered by this list the answer is not known. \[ \begin{aligned} (\Sigma _{1,4,5,2},(1,a))&=(\Sigma _{1,4,4},(0,a-2))~\ast ~ (\Sigma _{2,2},(1,2))~,&&a=6,~4~~~\, \mathrm{or}~~~\, 2~;\\ (\Sigma _{1,4,6,1},(1,a))&=(\Sigma _{1,4,4},(0,a-2))~\ast ~ (\Sigma _{3,1},(1,2))~,&&a=6,~4~~~\, \mathrm{or}~~~\, 2~;\\ (\Sigma _{1,4,4,3},(1,a))&=(\Sigma _{1,4,4},(0,a-2))~\ast ~ (\Sigma _{1,3},(1,2))~,&&a=6,~4~~~\, \mathrm{or}~~~\, 2~;\\ (\Sigma _{1,4,4,1,1,1},(1,a))&=(\Sigma _{1,4,4},(0,a-2))~\ast ~ (\Sigma _{1,1,1,1},(1,0))~,&&a=4~~~\, \mathrm{or}~~~\, 2~. \end{aligned} \] \hypertarget{(C2_1.2_theo9)}{1.2.} If now $\alpha _4=-$, the couple $\mathcal{K}_4$ is realizable, so one has to treat only the situation $\sigma ^4=\Sigma _{1,3,1}$ in which $\mathcal{K}^4$ is not realizable with the admissible pair $(0,2)$. This, combined with the list~\eqref{eqd=6}, gives the following sign patterns $\sigma _{\bullet}$: \[ \Sigma _{1,5,1,1,3,1}~,~~~\, \Sigma _{1,1,1,3,1,1,3,1}~,~~~\, \Sigma _{1,3,1,1,1,1,3,1}~,~~~\, \Sigma _{2,4,1,1,3,1}~~~\, \mathrm{and}~~~\, \Sigma _{1,4,2,1,3,1}~. \] \hypertarget{(C2_1.2.1_theo9)}{1.2.1.} The first, fourth and fifth sign patterns are realizable with the admissible pairs $(1,6)$, $(1,4)$ and $(1,2)$: \[ \begin{aligned} (\Sigma _{1,5,1,1,3,1},(1,a+b))&=(\Sigma _{1,4},(1,a))\ast (\Sigma _{2,1,1,3,1},(0,b))~,~~~a,b=1~~~\mathrm{or}~~~3~,\\ (\Sigma _{2,4,1,1,3,1},(1,a+b))&=(\Sigma _{2,3},(1,a))\ast \Sigma _{2,1,1,3,1},(0,b))~,~~~a,b=1~~~ \mathrm{or}~~~3~,\\ (\Sigma _{1,4,2,1,3,1},(1,a+b))&=(\Sigma _{1,4},(1,a))\ast (\Sigma _{1,2,1,3,1},(0,b))~,~~~a,b=1~~~ \mathrm{or}~~~3~. \end{aligned} \] \vspace*{1mm} \noindent\hypertarget{(C2_1.2.2_theo9)}{1.2.2.} The second (resp. the third) sign patterns is realizable with the admissible pairs $(1,4)$ and $(1,2)$ (resp. $(1,2)$); for the third sign pattern and for the admissible pair $(1,4)$, the answer is not known: \[ \begin{aligned} (\Sigma _{1,1,1,3,1,1,3,1},(1,4))&=(\Sigma _{1,1,1},(0,0))~\ast ~ (\Sigma _{1,3,1,1,3,1},(1,4))~,\\ (\Sigma _{1,1,1,3,1,1,3,1},(1,2))&=(\Sigma _{1,1,1,3,1,1,1},(0,0))~\ast ~ (\Sigma _{3,1},(1,2))~,\\ (\Sigma _{1,3,1,1,1,1,3,1},(1,2))&=(\Sigma _{1,3,1,1,1,1,1},(0,0))~\ast ~ (\Sigma _{3,1},(1,2))~. \end{aligned} \] \let\qed\relax \end{proof} \begin{proof}[\mdseries\hypertarget{(C2_2_theo9)}{2}] $\alpha _5=-$. The couple $\mathcal{K}_5$ is realizable while $\mathcal{K}^5$ is not only if $\sigma ^5=\Sigma _{1,4,1}$. \vspace*{1mm} \noindent \hypertarget{(C2_2.1_theo9)}{2.1.} If $\sigma ^5=\Sigma _{1,4,1}$ and $\alpha _6=+$, then $\mathcal{K}^6$ is realizable whereas $\mathcal{K}_6$ is not realizable only if $\sigma _6=\Sigma _{1,4,1}$, so $\sigma _{\bullet}=\Sigma _{1,4,1,1,4,1}$. This sign pattern is realizable with the admissible pairs $(1,4)$ and $(1,2)$ (for $(1,6)$, the answer remains unknown): \[ (\Sigma _{1,4,1,1,4,1},(1,4)~~~\, \mathrm{or}~~~\, (1,2))= (\Sigma _{1,3},(1,2))~\ast ~ (\Sigma _{2,1,1,4,1},(0,2)~~~\, \mathrm{or}~~~\, (0,0))~. \] \noindent \hypertarget{(C2_2.2_theo9)}{2.2.} If $\alpha _5=-$, $\sigma ^5=\Sigma _{1,4,1}$ and $\alpha _6=-$, i.e. $\sigma ^6=\Sigma _{2,4,1}$, then applying the involution $i_r$ (see~\eqref{eqir}) one transforms this case into $\sigma _5=\Sigma _{1,4,2}$, $\alpha _0=-$. This case was treated in~\hyperlink{(C2_1_theo9)}{1}. \let\qed\relax \end{proof} \let\qed\relax \end{proof} \let\qed\relax \end{proof} \begin{proof}[\hypertarget{(D_theo9)}{(D)}] Suppose that $d=12$. One can try to represent a given couple $\mathcal{K}_{\bullet}$ with $\tilde{\mu}\geq 2$ as a concatenation of couples $\mathcal{K}'$ and $\mathcal{K}''$ of degree $10$ and $2$ respectively. If $\mathcal{K}'$ is realizable, then such is $\mathcal{K}_{\bullet}$ as well, because all compatible couples of degree $2$ are realizable. So we assume that $\mathcal{K}'$ is not realizable. Then one can assume that $\mathcal{K}'$ is one of the two couples of part~(2) of the theorem. Hence the sign pattern of $\mathcal{K}'$ is among the following ones: \[ \begin{aligned} \Sigma _{1,4,5,3}~,~~ \Sigma _{1,4,5,2,1}~,~~\Sigma _{1,4,5,1,2}~,~~ \Sigma _{1,4,5,1,1,1}~,\\ \Sigma _{1,4,4,4}~,~~\Sigma _{1,4,4,3,1}~,~~ \Sigma _{1,4,4,2,2}~,~~\Sigma _{1,4,4,2,1,1}~. \end{aligned} \] Represent these sequences of $4$, $5$ or $6$ numbers in the forms $(1,A)$ and $(1,4,B)$, where $A$ is the sequence of the last $3$, $4$ or $5$ of them, and $B$ of the last $2$, $3$ or $4$ respectively. If $3\leq \pos \leq \negrm $, then \begin{equation}\label{eq12} \mathcal{K}_{\bullet}=(\Sigma _{1,1},(1,0))~\ast ~\mathcal{K}^{\sharp}~,~~~\, \mathrm{where}~~~\, \mathcal{K}^{\sharp}=(\Sigma _{1,A},(\pos -1,\negrm ))~. \end{equation} The couple $\mathcal{K}^{\sharp}$ is with $d=11$ and $\tilde{\mu}\geq 2$ hence realizable. Hence $\mathcal{K}_{\bullet}$ is also realizable. Suppose that $2=\pos \leq \negrm $. Then one can write \begin{equation}\label{eq12bis} \mathcal{K}_{\bullet}=(\Sigma _{1,4},(1,3)~~~\, \mathrm{or}~~~\, (1,1))~\ast ~ (\Sigma _{1,B},(a,b))~, \end{equation} where the admissible pair of $\mathcal{K}_{\bullet}$ is $(a+1,b+3)$ if $\negrm \geq 4$ or $(a+1,b+1)$ if $\negrm =2$ or~$3$. Each couple $(\Sigma _{1,4},(1,3))$, $(\Sigma _{1,4},(1,1))$ and $(\Sigma _{1,B},(a,b))$ is realizable (for the latter one has $d=8$ and $\tilde{\mu}\geq 1$), so $\mathcal{K}_{\bullet}$ is realizable. \let\qed\relax \end{proof} \begin{proof}[\hypertarget{(E_theo9)}{(E)}] Suppose that $d=13$ or $d=14$. (For $d=14$, we assume that the theorem is proved for $d=13$.) Similarly to part~\hyperlink{(D_theo9)}{(D)}, one can try to represent a given couple $\mathcal{K}_{\bullet}$ with $\tilde{\mu}\geq 2$ as a concatenation of couples $\mathcal{K}'$ and $\mathcal{K}''$ of degree $10$ and $3$ respectively (or $10$ and $4$ if $d=14$). As in part~\hyperlink{(D_theo9)}{(D)}, we assume that $\mathcal{K}'$ is one of the couples of part~(2) of the theorem. We denote the sign pattern of $\mathcal{K}_{\bullet}$ by $(1,A)$ and $(1,4,B)$ with similar definition of $A$ and $B$. For $3\leq \pos \leq \negrm $, we use formula~\eqref{eq12}. The couple $\mathcal{K}^{\sharp}$ is with $d=12$ (resp. $d=13$) and $\tilde{\mu}\geq 2$ hence realizable. For $2=\pos \leq \negrm $, we can use formula~\eqref{eq12bis} with the reasoning after it except in the case when $(\Sigma _{1,B},(a,b))$ is of the orbit of the couple~\eqref{eqcased9} (resp. of one of the orbits of the couples of part~(2) of the theorem). The first $8$ signs of the sign pattern of $\mathcal{K}_{\bullet}$ are $(+,-,-,-,-,+,+,+)$. Further we treat separately the cases $d=13$ and $d=14$. Suppose that $d=13$. These first $8$ signs imply that from the orbit of the sign pattern $\Sigma _{1,4,4,1}=i_r(\Sigma _{1,4,4,1})$ one has to choose $\Sigma _{1,4,4,1}$ (and not $\Sigma _{2,1,1,2,1,1,2}=i_m(\Sigma _{1,4,4,1})=i_r(\Sigma _{2,1,1,2,1,1,2})$) to be equal to $\Sigma _{1,B}$. Thus the sign pattern of $\mathcal{K}_{\bullet}$ is $\Sigma _{1,4,4,4,1}$. The admissible pair equals $(2,\nu )$, $\nu =9$, $7$, $5$ or $3$. We set \[ (\Sigma _{1,4,4,4,1},(2,\nu ))=(\Sigma _{1,2},(1,1))~\ast ~ \mathcal{K}^*~\ast ~(\Sigma _{1,1},(1,0))~,~~~\, \mathcal{K}^*\coloneqq (\Sigma _{3,4,4},(0,\nu -1)) \] The first and the third couple in this concatenation are realizable. The couple $\mathcal{K}^*$ is realizable for $\nu =9$, see \cite[Theorem~4]{CGK1}. Hence there exists a real polynomial $Q_{\natural}$ with sign pattern $\Sigma _{3,4,4}$ and having $8$ negative distinct roots and a complex conjugate pair. One can perturb the coefficients of $Q_{\natural}$ without changing its sign pattern so that all its critical values become distinct. Hence for suitable positive values of $t$ one obtains polynomials $Q_{\natural}+t$ having the sign pattern $\Sigma _{3,4,4}$ and with exactly $6$, $4$ or $2$ distinct negative roots and, respectively, $2$, $3$ or $4$ conjugate pairs. Thus the couple $\mathcal{K}^*$ (and $(\Sigma _{1,4,4,4,1},(2,\nu ))$ as well) is realizable for $\nu =9$, $7$, $5$ or $3$. \let\qed\relax \end{proof} \begin{proof}[\hypertarget{(F_theo9)}{(F)}] Suppose that $d=14$. The first $8$ signs of the sign pattern of $\mathcal{K}_{\bullet}$ (see~\hyperlink{(E_theo9)}{(E)}) mean that the sign pattern $\Sigma _{1,B}$ is in one of the orbits of the sign patterns $\Sigma _{1,4,5,1}$ or $\Sigma _{1,4,4,2}$, see part~(2) of the theorem; the first component of $B$ must be $\geq 3$. Hence $\Sigma _{1,B}$ is among the following sign patterns: \[ \Sigma _{1,4,5,1}~,~~~\, \Sigma _{1,5,4,1}~~~\, \mathrm{or}~~~\, \Sigma _{1,4,4,2}~. \] For $\Sigma _{1,B}=\Sigma _{1,4,4,2}$, one has \[ \mathcal{K}_{\bullet}=(\Sigma _{1,4,4,4,2},(2,\nu +1))= (\Sigma _{1,4,4,4,1},(2,\nu ))~\ast ~(\Sigma _{2},(0,1))~, \] where $\nu =9$, $7$, $5$ or $3$ and the first concatenation factor is realizable (see~\hyperlink{(E_theo9)}{(E)}). Hence $\mathcal{K}_{\bullet}$ is also realizable. For $\Sigma _{1,B}=\Sigma _{1,4,5,1}$, one has \[ \mathcal{K}_{\bullet}=(\Sigma _{1,4,4,5,1},(2,\nu +1))=(\Sigma _{1,2},(1,1)) ~\ast ~\mathcal{K}^*~\ast ~(\Sigma _{2,1},(1,1)) \] with $\mathcal{K}^*$ as in~\hyperlink{(E_theo9)}{(E)}, so this case is also realizable. For $\Sigma _{1,B}=\Sigma _{1,4,5,1}$, the sign pattern of $\mathcal{K}_{\bullet}$ equals $\Sigma _{1,4,5,4,1}$. It is realizable with the admissible pairs $(2,\rho )$, $\rho =2$, $4$, $6$ or $8$: \[ (\Sigma _{1,4,5,4,1},(2,\rho ))=(\Sigma _{1,4,1},(0,1))~\ast ~ (\Sigma _{5,4,1},(2,\rho -1))~. \] For $\rho =10$, the answer to Problem~\ref{prob1} remains unknown. \let\qed\relax \end{proof} \section{Proof of Theorem~\texorpdfstring{\protect\ref{tm2SC}}{10}}\label{secprtm2SC} \begin{proof}[\hypertarget{(A_theo10)}{(A)}] We consider first the (non)-realizable cases with the admissible pairs $(0,1)$ and $(0,3)$. Every couple $(\Sigma _{m,n,q},(0,1))$ with $m+n+q=2\ell$, $\ell \geq 2$, $m>0$, $n>0$, $q>0$, is realizable -- it suffices to choose a polynomial with sign pattern $\Sigma _{m,n,q}$ and to add to it a large positive constant. Further we suppose that $q\geq m$, otherwise one considers the couple from the same orbit $(\Sigma _{q,n,m},(0,3))$ using the involution~$i_r$. On the other hand any couple of the form $(\Sigma _{d+1},(0,k))$ is realizable (\cite[Example~1]{CGK1}). This implies that for $d=9$, any couple with $q\geq 3$ and admissible pair $(0,3)$ is realizable as \[ (\Sigma _{m,n,q},(0,3))=(\Sigma _{m,n,q-2},(0,1))\ast (\Sigma _3,(0,2))~.\] The couple $(\Sigma _{2,6,2},(0,3))$ is realizable as \[ (\Sigma _{2,6,2},(0,3))=(\Sigma _{2,6,1},(0,2))\ast (\Sigma _2,(0,1))~, \] where the first factor is from the same orbit as \[ (\Sigma _{1,6,2},(0,2))=(\Sigma _{1,6,1},(0,1))\ast (\Sigma _2,(0,1)) \] hence this orbit is realizable. We prove realizability of the orbit $(\Sigma _{1,7,2},(0,3))$ by direct construction. We set \[ G_0\coloneqq x(x^2-1)^2(x^4+2x^2+1)=x^9-2x^5+x~, \] and then $G_1\coloneqq G_0-0.0001(x+1)^8+0.2$ which has three negative roots $-1.09\ldots$, $-0.84\ldots$ and $-0.20\ldots$ and three complex conjugate pairs of roots. The sign pattern defined by $G_1$ is $\Sigma _{1,7,2}$, because \begin{multline*} G_1=x^9-0.0001x^8-0.0008x^7-0.0028x^6-2.0056x^5-0.0070x^4 \\ -0.0056x^3-0.0028x^2+ 0.9992x+0.1999~. \end{multline*} Finally, the orbit $(\Sigma _{1,8,1},(0,3))$ is not realizable, see \cite[part (i) of Theorem~4]{KoSh1} (one has to apply the involution $i_m$ to the series of non-realizable examples described there). \let\qed\relax \end{proof} \begin{proof}[\hypertarget{(B_theo10)}{(B)}] Suppose that the admissible pair is $(0,5)$. If $q\geq 5$, then one realizes the couple by the concatenation \[ (\Sigma _{m,n,q-4},(0,1))\ast (\Sigma _5,(0,4))~. \] Suppose that either $q=4$ or $q=3$ and $m=2$ or $3$. Then by \cite[Theorem~9]{FoKoSh} the corresponding couple $(\Sigma _{m,n,q-2},(0,3))$ is realizable and one sets \[ (\Sigma _{m,n,q},(0,5))=(\Sigma _{m,n,q-2},(0,3))\ast (\Sigma _3,(0,2))~.\] The couple $(\Sigma _{1,8,1},(0,5))$ is not realizable (see \cite[part (i) of Theorem~4]{KoSh1}). The following proposition settles the remaining cases: \begin{prop}\label{prop3cases} The couples $(\Sigma _{1,7,2},(0,5))$, $(\Sigma _{1,6,3},(0,5))$ and $(\Sigma _{2,6,2},(0,5))$ are not realizable. \end{prop} The proposition is proved in Section~\ref{secprprop3cases}. \let\qed\relax \end{proof} \begin{proof}[\hypertarget{(C_theo10)}{(C)}] Suppose that $d=9$ and the admissible pair is $(0,7)$. The couple $(\Sigma _{1,8,1},(0,7))$ is not realizable, see \cite[part (i) of Theorem~4]{KoSh1}. It follows from \cite[Theorem~3]{CGK1} that if $n=1$, $2$ or $3$, then such a sign pattern is realizable with the admissible pair $(0,7)$. The same theorem implies that for $m=1$, $n\geq 4$, the couple $(\Sigma _{m,n,q},(0,7))$ is not realizable. Proposition~1 in \cite{CGK1} says that the couples $(\Sigma _{3,4,3},(0,7))$ and $(\Sigma _{2,4,4},(0,7))$ are not realizable. The remaining couples $(\Sigma _{2,5,3},(0,7))$ and $(\Sigma _{2,6,2},(0,7))$ are not realizable by \cite[Proposition~6]{FoKoSh}. \let\qed\relax \end{proof} \section{Proof of Proposition\texorpdfstring{\protect\ref{prop3cases}}{19}}\protect\label{secprprop3cases} \begin{defi}\ \begin{enumerate} \item\label{prop20_1} A \emph{generalized sign pattern} is a string of signs $+$, $-$ and/or $0$ beginning with a $+$. A real polynomial $P$ with positive leading coefficient is said to define a given generalized sign pattern $\sigma$ if the components of $\sigma$ are equal to the signs of the corresponding coefficients of $P$. Given a sign pattern $\sigma _1$ and a generalized sign pattern $\sigma _2$ of the same length, $\sigma _2$ is called \emph{adjacent} to $\sigma _1$ if it is obtained from $\sigma _1$ by replacing some of its components (excluding the initial $+$) by zeros. The \emph{closure} of a given sign pattern is the set containing the sign pattern and all generalized sign patterns adjacent to it. \item\label{prop20_2} We call \emph{simultaneous shift} a map $\tau _{\pm}:x\mapsto x\pm \varepsilon$, where $\varepsilon >0$. We usually consider $\varepsilon$ to be sufficiently small. \end{enumerate} \end{defi} \begin{lemma}\label{lmshift} Given a real monic polynomial $W\coloneqq \sum _{j=0}^da_jx^j$ with at least one vanishing coefficient, there exists a simultaneous shift $\tau _{\pm}$ after which all coefficients of $W$ are non-zero and the initially non-zero coefficients keep their signs. Moreover, if $a_k=0\neq a_{k+1}$, then one can choose the sign $+$ or $-$ in the definition of $\tau _{\pm}$ so that after the shift the sign of the new coefficient $a_k$ be the desired one ($+$ or $-$). \end{lemma} \begin{proof} After the shift all coefficients of $W$ become non-constant polynomials in $\varepsilon$, so with the exception of finitely-many values of $\varepsilon$ they are all non-zero. After the shift the coefficient $a_k$ becomes $a_k\pm (k+1)a_{k+1}\varepsilon +o(\varepsilon )$ from which the last statement of the lemma follows. \end{proof} \begin{lemma}\label{lmnotexist7} For $d=7$, there exists no monic polynomial $P\coloneqq \sum _{j=0}^7a_jx^j$ satisfying simultaneously the following conditions: \begin{enumerate} \item\label{lemma22_i} one has $P(0)>0$ and $P$ defines either one of the sign patterns $\Sigma _{1,6,1}$ or $\Sigma _{1,5,2}$ or a generalized sign pattern adjacent to one of them; \item\label{lemma22_ii} $P$ has $5$ negative roots counted with multiplicity; \item\label{lemma22_iii} $P$ has either a complex conjugate pair of roots or a double positive root. \end{enumerate} \end{lemma} \begin{proof} Suppose first that $P$ has a complex conjugate pair of roots. Denote by $-\eta _j$ the negative roots of $P$. Set $P_1\coloneqq x(x+\eta _1)\cdots (x+\eta _5)$. Hence all coefficients of $P_1$ are positive. For $\varepsilon >0$ small enough, the polynomial $P_2\coloneqq P-\varepsilon P_1$ defines one of the sign patterns $\Sigma _{1,6,1}$ or $\Sigma _{1,5,2}$ and has a complex conjugate pair. One can perturb the negative roots of $P_2$ to make them all distinct while keeping the signs of its coefficients the same and the presence of a complex conjugate pair. However such a polynomial $P_2$ does not exist, see \cite[Theorem~9]{FoKoSh}. Suppose that $P$ has a double positive root. Hence $P$ is hyperbolic. If $P$ has no vanishing coefficients, then one can perturb the negative roots of $P$ to make them all distinct without changing the signs of the coefficients. After this one considers the polynomial $P_3\coloneqq P+\delta x^4$ with $5$ distinct negative roots and defining the same sign pattern as $P$ when $\delta >0$ is small enough. The polynomial $P_3$ has no double positive root, but a complex conjugate pair close to the double root of $P$. Again by \cite[Theorem~9]{FoKoSh} such a polynomial $P_3$ does not exist. Suppose that $P$ has a double positive root and at least one vanishing coefficient. We remind that $P$ cannot have two consecutive vanishing coefficients (\cite[Lemma 7]{KoMB}). For $\varepsilon >0$ small enough, all coefficients of $P$ can be made non-zero as a result of a shift $\tau _{\pm}$. We consider the following cases: \begin{proof}[Case \hypertarget{proof_lemma22_c1}{1}] One has $a_6<0$ and $a_1<0$ (resp. $a_2<0$). A shift $\tau_+$ or $\tau _-$ makes all coefficients between $a_6$ and $a_1$ (resp. between $a_6$ and $a_2$) non-zero. Hence they are all negative, otherwise by Descartes' rule of signs $P$ cannot have $5$ negative roots. One perturbs the negative roots of $P$ to make them distinct while keeping the sign pattern. Then for $\delta >0$ small enough, the polynomial $P+\delta x^4$ has still $5$ distinct negative roots and the same sign pattern, but the double root gives birth to a complex conjugate pair close to it. Such a polynomial does not exist, see \cite[Theorem~9]{FoKoSh}. \let\qed\relax \end{proof} \begin{proof}[Case \hypertarget{proof_lemma22_c2}{2}] One has $a_6=0$ and $a_1<0$ (resp. $a_2<0$). A shift $\tau _{\pm}$ with suitably chosen sign $+$ or $-$ makes all coefficients between $a_7$ and $a_1$ (resp. between $a_7$ and $a_2$) non-zero, and in particular $a_6$ becomes negative. Again by Descartes' rule of signs the rest of the coefficients between $a_7$ and $a_1$ (resp. between $a_7$ and $a_2$) are negative and as in case~\hyperlink{proof_lemma22_c1}{1} one concludes that such a polynomial does not exist. \let\qed\relax \end{proof} \begin{proof}[Case \hypertarget{proof_lemma22_c3}{3}] One has $a_6<0$ and $a_1=0$ (resp. $a_2=0$). A shift $\tau _{\pm}$ with suitably chosen sign $+$ or $-$ makes all coefficients between $a_6$ and $a_0$ (resp. between $a_6$ and $a_1$) non-zero, and in particular $a_1$ (resp.~$a_2$) becomes negative. The rest of the reasoning is as in case~\hyperlink{proof_lemma22_c1}{2}. \let\qed\relax \end{proof} \begin{proof}[Case \hypertarget{proof_lemma22_c4}{4}] One has $a_6=a_1=0$ (resp. $a_6=a_2=0$). Then $a_2\neq 0$ (resp. $a_3\neq 0$), see \cite[Lemma 7]{KoMB}, hence $a_2<0$ (resp. $a_3<0$). A shift $\tau _{\pm}$ with suitably chosen sign $+$ or $-$ makes all coefficients between $a_7$ and $a_0$ (resp. between $a_7$ and $a_1$) non-zero, and in particular $a_6$ becomes negative. Hence the sign pattern of $P$ is now $\Sigma _{1,6,1}$, $\Sigma _{1,5,2}$ or $\Sigma _{1,4,3}$. One perturbs the negative roots of $P$ to make them distinct while preserving the sign pattern, and then adds to $P$ the monomial $\delta x^4$ with $\delta >0$ small enough. The double root gives birth to a complex conjugate pair and $P$ realizes one of the couples $(\Sigma _{1,k,7-k},(0,5))$, $k=4$, $5$ or~$6$, which by \cite[Theorem~9]{FoKoSh} is impossible. \end{proof} \let\qed\relax \end{proof} \goodbreak \begin{lemma}\label{lmnotexist9} There exists no real monic degree $9$ polynomial $U$ satisfying the following conditions: \begin{enumerate} \item\label{lemma23_i} %(i) its sign pattern is $\Sigma _{1,7,2}$, $\Sigma _{1,6,3}$ or $\Sigma _{2,6,2}$ or it is a generalized sign pattern adjacent to one of these sign patterns; \item\label{lemma23_ii} % (ii) it has $7$ negative roots counted with multiplicity; \item\label{lemma23_iii} % (iii) it has either a double positive root or a complex conjugate pair. \end{enumerate} \end{lemma} \begin{proof} First of all we observe that the last three components of the (generalized) sign pattern cannot be $(+,0,+)$, because in this case the polynomial $U(-x)$ has less than $7$ sign changes and by Descartes' rule of signs the polynomial cannot have $7$ negative roots. Hence the coefficients of $x^8$, $\ldots$, $x^3$ (or $x^7$, $\ldots$, $x^3$ in the case of $\Sigma _{2,6,2}$) are non-positive. The one of $x$ is non-negative and can be $0$ only if the one of $x^2$ is non-positive. If the polynomial $U$ has a double positive root, then it is hyperbolic. Recall that by \cite[Lemma 7]{KoMB} the polynomial $U$ has no two consecutive vanishing coefficients. One can perform a simultaneous shift to make all coefficients non-zero and so that the coefficient of $x^8$ (or of $x^7$ in the case of $\Sigma _{2,6,2}$) is negative. By Descartes' rule of signs the sign pattern of the polynomial is now $\Sigma _{1,5,4}$, $\Sigma _{1,6,3}$, $\Sigma _{1,7,2}$, $\Sigma _{2,6,2}$ or $\Sigma _{2,7,1}$. One can perturb its negative roots to make them distinct and then add to the polynomial a monomial $\delta x^4$, where $\delta >0$ is small enough. Thus the sign pattern is preserved and the double positive root gives birth to a complex conjugate pair close to it. However such a polynomial does not exist, see \cite[Proposition~6]{FoKoSh}. If $U$ has a complex conjugate pair, then we change it to $U_1\coloneqq U-\varepsilon \prod _{j=1}^7(x+\eta _j)$, $-\eta _j$ being the negative roots of $U$. For $\varepsilon >0$ small enough, all coefficients of $U_1$ with the possible exception only of the one of $x^8$ are non-zero. After a simultaneous shift the coefficient of $x^8$ becomes negative and the sign pattern now equals $\Sigma _{1,6,3}$, $\Sigma _{1,7,2}$, $\Sigma _{1,8,1}$ or $\Sigma _{2,6,2}$. Then one perturbs the negative roots to make them distinct, so the polynomial realizes one of the couples $(\Sigma _{1,6,3},(0,7))$, $(\Sigma _{1,7,2},(0,7))$, $(\Sigma _{1,8,1},(0,7))$ or $(\Sigma _{2,6,2},(0,7))$ which is impossible, see \cite[Proposition~6]{FoKoSh}. \end{proof} \begin{defi} We call \emph{multiplicity vector} the vector whose components are equal to the multiplicities of the negative roots of a real polynomial listed in the increasing order. The multiplicity vector $\vec{v}_1$ is \emph{adjacent} to the multiplicity vector $\vec{v}_2$ if $\vec{v}_1$ is obtained by applying one or several times the operation of replacing two consecutive components by their sum. Example: $(5)$ is adjacent to $(3,2)$ which in turn is adjacent to $(3,1,1)$. \end{defi} \begin{lemma}\label{lmtransform} Suppose that there exists a degree $9$ real monic polynomial $V$ realizing the couple $(\Sigma _{1,7,2},(0,5))$ (resp.~$(\Sigma _{1,6,3},(0,5))$ or $(\Sigma _{2,6,2},(0,5))$). Then there exists a real monic degree $9$ polynomial satisfying the following conditions: \begin{enumerate} \item\label{lemma25_i} % (i) it defines the sign pattern $\Sigma _{1,7,2}$ (resp. $\Sigma _{1,6,3}$ or $\Sigma _{2,6,2}$); \item\label{lemma25_ii} % (ii) it has a double positive root; \item\label{lemma25_ii} % (iii) the multiplicity vector of its negative roots is among the following ones: \[ (1,2,2)~,~~~\, (2,1,2)~,~~~\, (3,1,1)~,~~~\, (1,1,3)~,~~~\, (3,2)~, ~~~\, (2,3)~. \] \end{enumerate} \end{lemma} \begin{proof} The polynomial $V$ has either $4$ or $6$ critical points for $x<0$ (counted with multiplicity). We denote them by $-\xi _4<-\xi _3<-\xi _2<-\xi _1$; if they are $6$, then between two consecutive negative roots of $V$ there are three critical points of which we choose the rightmost one. We denote the corresponding critical values by $\eta _i$, where $\eta _4>0$, $\eta _3<0$, $\eta _2>0$ and $\eta _1<0$. Suppose that $\eta _3=\eta _1$. Then we add to $V$ a positive constant (this does not change the sign pattern) to obtain a polynomial with multiplicity vector $(1,2,2)$. Suppose that $\eta _3<\eta _1$. We consider the family of polynomials $V_t\coloneqq V+tx$, $t\geq 0$ in which the sign of the coefficient of $x$ is $+$. As $t$ increases, the critical value $\eta _4$ decreases faster than $\eta _3$ and $\eta _2$ decreases faster than $\eta _1$. Denote by $t_0>0$ the smallest value of $t$ for which one of the following things happens: % \vspace{1mm} \begin{enumerate} \item\label{proof_lemma25_1} % 1) One has $\eta _4=\eta _1$. In this case we add to $V_{t_0}$ a positive constant to obtain a polynomial with multiplicity vector $(2,1,2)$. % \vspace{1mm} \item\label{proof_lemma25_2} % 2) One has $\eta _2=\eta _1$. Then $-\xi _1$ is a degenerate critical point of $V_{t_0}$. We add to $V_{t_0}$ a positive constant and get a polynomial with multiplicity vector $(1,1,3)$. % \vspace{1mm} Suppose that $\eta _3>\eta _1$. Then for $t=t_0$, one of the following things can take place in the family $V_t$: % \vspace{1mm} \item\label{proof_lemma25_3} % 3) One has $\eta_3=\eta _1$. We add to $V_{t_0}$ a positive constant and obtain a polynomial with multiplicity vector $(1,2,2)$. % \vspace{1mm} \item\label{proof_lemma25_4} % 4) One has $\eta _4=\eta _3$. Then $-\xi _3$ is a degenerate critical point of $V_{t_0}$. We add to $V_{t_0}$ a positive constant and get a polynomial with multiplicity vector $(3,1,1)$. % \vspace{1mm} \end{enumerate} If~\eqref{proof_lemma25_1} and~\eqref{proof_lemma25_2} (resp.~\eqref{proof_lemma25_3} and~\eqref{proof_lemma25_4}) take place simultaneously, then the multiplicity vector of $V_{t_0}$ is $(2,3)$ (resp. $(3,2)$). It is not possible for $t=t_0$ to obtain an equality between $\eta _1$, $\eta _2$, $\eta _3$ or $\eta _4$ and one of the two possible other critical values of $V_{t_0}$ (if $V_{t_0}$ has $6$ and not $4$ negative critical points), because then one can add a positive constant to $V_{t_0}$ and get a polynomial with $7$ negative roots, one conjugate pair and sign pattern $\Sigma _{1,7,2}$, $\Sigma _{1,6,3}$ or $\Sigma _{2,6,2}$ which by Lemma~\ref{lmnotexist9} is impossible. \end{proof} \begin{prop} There exists no real monic degree $9$ polynomial having $5$ negative roots with multiplicity vector having $1$, $2$ or $3$ components, from the closure of the sign pattern $\Sigma _{1,7,2}$, $\Sigma _{1,6,3}$ or $\Sigma _{2,6,2}$, and having one double positive root. \end{prop} \begin{proof}\ \relax \begin{proof}[\hypertarget{prop26_A}{(A)}] We explain the method of proof on the example of the multiplicity vector $(5)$ (one five-fold negative root). We want to prove the non-existence of a polynomial \[ H_5\coloneqq (x+1)^5((x+u)^2+v)(x-c)^2~,~~~\, v>0~,~~~\, c>0~,~~~\, u\in \mathbb{R}~. \] One rescales the $x$-axis to make the negative root equal to $-1$. The index $5$ corresponds to the multiplicity vector $(5)$. Hence the non-existence of such a polynomial $H_5\coloneqq \sum _{j=0}^9a_jx^j$ is tantamount to the emptyness of the domain $E_5\cap D_5$, where $D_5\subset \mathbb{R}^3$ is defined by the inequalities \[ D_5~:~\{ ~(u, v, c)~|~v>0~,~c>0~\} \] and $E_5$ is the closed domain in $\mathbb{R}^3$ defined by the condition the signs of the coefficients of $H_5$ to correspond to the closure of one of the sign patterns $\Sigma _{1,7,2}$, $\Sigma _{1,6,3}$ or $\Sigma _{2,6,2}$. We remind that each coefficient of $H_5$ is a polynomial in the variables $(u,v,c)$, with $a_0=c^2(v+u^2)$, $\ldots$, $a_8=5+2u-2c$. If the interior of the domain $E_5$ is non-empty, then $E_5$ has a priori the structure of a stratified manifold. Its stratum of maximal dimension corresponds to polynomials defining the sign pattern $\Sigma _{1,7,2}$, $\Sigma _{1,6,3}$ or $\Sigma _{2,6,2}$. The closure $\bar{D}_5$ of the set $D_5$ is defined by the inequalities $v\geq 0$, $c\geq 0$ and its border $\partial D_5$ by $v\geq 0$, $c\geq 0$, $cv=0$. No polynomial $H_5\in \partial D_5$ is from the closure of the sign pattern $\Sigma _{1,7,2}$, $\Sigma _{1,6,3}$ or $\Sigma _{2,6,2}$. Indeed, for $c=00=u=v$, one has $H_5=x^2H^1_5$, where $H^1_5(0)>0$. Then the (generalized) sign pattern of $H_5$ cannot be adjacent to $\Sigma _{1,7,2}$ or $\Sigma _{2,6,2}$, but only to $\Sigma _{1,6,3}$ and the one of $H^1_5$ is adjacent to $\Sigma _{1,6,1}$. By Lemma~\ref{lmnotexist7} such a polynomial $H^1_5$ does not exist. For $v=00$, non-existence of $H_5$ follows from Lemma~\ref{lmnotexist9}. For $v=00$. At the point $(0,0,0)$ at least one coefficient $a_j$ of $H_5$ has the wrong sign. The least possible value of $r_0$ is the least one for which $H_5\in E_5$, i.e. where the signs of all coefficients correspond to the closure of the sign pattern $\Sigma _{1,7,2}$, $\Sigma _{1,6,3}$ or $\Sigma _{2,6,2}$. So for $r=r_0$, at least one of the coefficients $a_j$ of $H_5$ vanishes, because the corresponding polynomial(s) $H_5$ belong to the border, but not to the interior of the set~$E_5$. We use the method of Lagrange's multipliers as follows. We are looking for the minimal value of the function $T_0^{\pm}$ on the hypersurface $\{ a_j=0\}$. We construct the function \[ \tilde{T}^{\pm}_j\coloneqq \pm u+v+c+\lambda a_j~, \] where $\lambda$ is a Lagrange multiplier. For $j=0$, $\ldots$, $8$, we consider the system of equations \[ \partial \tilde{T}^{\pm}_j/\partial \lambda = a_j=\partial \tilde{T}^{\pm}_j/\partial u= \partial \tilde{T}^{\pm}_j/\partial v= \partial \tilde{T}^{\pm}_j/\partial c=0~. \] In each case we are looking for a solution with $\lambda \in \mathbb{R}$ and $(u,v,c)\in D_5^{\pm}$. It turns out that in each case either there is no solution or the sign of $u$, $v$ or $c$ of the solution is not the right one or the signs of the coefficients of the obtained polynomial $H_5$ are in contradiction with the closure of the sign pattern $\Sigma _{1,7,2}$, $\Sigma _{1,6,3}$ or $\Sigma _{2,6,2}$. This can be established using computer algebra. \let\qed\relax \end{proof} \begin{proof}[\hypertarget{prop26_C}{(C)}] The multiplicity vector $(5)$ is adjacent to four multiplicity vectors with two components: $(4,1)$, $(3,2)$, $(2,3)$ and $(1,4)$. We prove the non-existence of polynomials \[ H_{k,5-k}\coloneqq (x+\mu )^k(x+1)^{5-k}((x+u)^2+v)(x-c)^2~,~~~\, k=1,~2,~3,~4~, \] with $\mu >1$ and with $(u,v,c)$ as above. Hence \[ \mathbb{R}^4\supset D_{k,5-k}=\{ (u,v,c,\mu )~|~v>0~,~c>0~,~\mu >1~\} ~. \] \looseness-1 As in~\hyperlink{prop26_A}{(A)} one shows that no polynomial $H_{k,5-k}\in \partial D_{k,5-k}$ with $c=0$ or $v=0$ has signs of the coefficients from the closure of $\Sigma _{1,7,2}$, $\Sigma _{1,6,3}$ or $\Sigma _{2,6,2}$. For $\mu =1$, one is looking in fact for a polynomial $H_5$ about which it was shown in~\hyperlink{prop26_A}{(A)}--\hyperlink{prop26_B}{(B)} that it does not exist. Hence $E_{k,5-k}\cap \partial D_{k,5-k}=\emptyset$. \let\qed\relax \end{proof} \begin{proof}[\hypertarget{prop26_D}{(D)}] To prove that $E_{j,5-j}\cap D_{j,5-j}=\emptyset$ we use again the method of Lagrange multipliers. We set $D_{k,5-k}^+\coloneqq D_{k,5-k}\cap \{ u>0\}$, $D_{k,5-k}^-\coloneqq D_{k,5-k}\cap \{ u<0\}$, $S_r^{\pm}:\pm u+v+c+(\mu -1)=r$ ($r\in \mathbb{R}$) and $\tilde{S}_j^{\pm}\coloneqq \pm u+v+c+(\mu -1)+\lambda a_j$. In this case \[ a_8=j\mu +(5-j)+2u-2c~,~\ldots ~,~a_0=\mu ^jc^2(v+u^2)~. \] For $j=0$, $\ldots$, $8$, and for $k$ as above, we consider the system of equations \[ \partial \tilde{S}^{\pm}_j/\partial \lambda =a_j= \partial \tilde{S}^{\pm}_j/\partial u= \partial \tilde{S}^{\pm}_j/\partial v= \partial \tilde{S}^{\pm}_j/\partial c= \partial \tilde{S}^{\pm}_j/\partial \mu =0~. \] We are looking for a solution with $\lambda \in \mathbb{R}$ and $(u,v,c,\mu )\in D_{k,5-k}^{\pm}$. In each case either there is no real solution or the sign of $u$, $v$, $c$ or $\mu -1$ of the solution is not the right one. \let\qed\relax \end{proof} \begin{proof}[\hypertarget{prop26_E}{(E)}] The possible multiplicity vectors with three components are $(2,2,1)$, $(2,1,2)$, $(1,2,2)$, $(3,1,1)$, $(1,3,1)$ and $(1,1,3)$. The polynomials $H_{j,k,5-j-k}$ defined after the multiplicity vectors $(j,k,5-j-k)$ are \[ H_{j,k,5-j-k}\coloneqq (x+\mu _2)^j(x+\mu _1)^k(x+1)^{5-j-k}((x+u)^2+v)(x-c)^2~, \] with $(u,v,c)$ as above and $1<\mu _1<\mu _2$. We set \[ \mathbb{R}^5\supset D_{j,k,5-j-k}\coloneqq \{ (u,v,c,\mu _1,\mu _2)~|~v>0~,~c>0~,~ 1<\mu _1~,1<\mu _2~\} ~, \] i.e.~we consider a domain larger than the strict analog of the domains $D_5$ and $D_{j,5-j}$. (The strict analog would be defined by $1<\mu _1<\mu _2$ instead of $1<\mu _1$, $1<\mu _2$.) This is done with the aim to simplify the computations. We set also $D_{j,k,5-j-k}^-\coloneqq D_{j,k,5-j-k}\cap \{ u<0\}$, $K_r^{\pm}:\pm u+v+c+(\mu _1-1)=r$ ($r\in \mathbb{R}$) and $\tilde{K}_i^{\pm}\coloneqq \pm u+v+c+(\mu -1)+(\mu _2-1)+\lambda a_i$. For $i=0$, $\ldots$, $8$, and for $(j,k)$ as above, we consider the system of equations \[ \partial \tilde{K}^{\pm}_i/\partial \lambda =a_i=\partial \tilde{K}^{\pm}_i/\partial u= \partial \tilde{K}^{\pm}_i/\partial v= \partial \tilde{K}^{\pm}_i/\partial c= \partial \tilde{K}^{\pm}_i/\partial \mu _1= \partial \tilde{K}^{\pm}_i/\partial \mu _2=0~. \] We are looking for a solution with $\lambda \in \mathbb{R}$ and $(u,v,c,\mu_1,\mu _2 )\in D_{j,k,5-j-k}^{\pm}$. In each case either there is no real solution or the sign of $u$, $v$, $c$, $\mu _1-1$ or $\mu _2-1$ of the solution is not the right one or, when a solution exists, the obtained polynomial $H_{j,k,5-j-k}$ does not have a (generalized) sign pattern from the closure of $\Sigma _{1,7,2}$, $\Sigma _{1,6,3}$ or $\Sigma _{2,6,2}$. \end{proof} \let\qed\relax \end{proof} \section*{Declaration of interests} The authors do not work for, advise, own shares in, or receive funds from any organization that could benefit from this article, and have declared no affiliations other than their research organizations. \printbibliography \end{document}