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Such a polynomial may be represented as $q^2 +xr^2$ (proof: this is true for all irreducible divisors which have degree 1 or 2, and the set of polynomials of this kind is a multiplicative monoid by Euler identity.) Strictly positive for $x>0$ polynomials may be represented as $q(x)/(x+1)^n$, where $q$ has non-negative coefficients.

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You can parametrize the zero locus of $\Delta$ (other than the origin) in the first quadrant by $$(x,y) = \left(\frac{(3{-}t)\sqrt{(3{+}t)(1{-}t)}}{8}, \frac{(3{+}t)\sqrt{(3{-}t)(1{+}t)}}{8}\right) \qquad\qquad -1\le t\le 1$$ and then compute that the curvature is $$\frac{(x'y''-y'x'')}{((x')^2+(y')^2)^{3/2}} = \frac{2^{3/2}(3{+}t^2)}{\... 11 The most abstract version of the Tarski-Seidenberg theorem I know of is the following Let f:A\to B be a morphism of finite presentation of commutative rings. Then the induced map$$f^*:\operatorname{Sper}B\to \operatorname{Sper}A$$sends constructible sets to constructible sets. Here \operatorname{Sper} is the real spectrum of the ring, i.e. the set of ... 6 Overview: The boundaries of the basins of attraction are lower dimensional stable manifolds. In two dimensions, they are the arcs flowing from repelling fixed points to saddle points. I expect that these are basically never analytic near the repelling fixed point. More specifically, if f(x,y) = \alpha x^2 + \beta y^2 + (\mbox{higher order terms}), then dx/... 6 There are many interesting examples where there are singularities on the border. E.g. spectrahedra describing decompositions of polynomials as sums of squares (s.o.s.) of polynomials have singularities corresponding to minimal possible number of squares in such decompositions. Specifically, if f(x_1,...,x_n)=\sum_k q_k(x_1,...,x_n)^2 then f=X^\top A X,... 4 The answer to this question is negative: Jut take \alpha such that 2^\alpha=3. Taking into account that 2^n\ne 3^m for any natural numbers n,m, we can prove that the number \alpha=\log_23 is irrational. Now observe that the set A=\{(x,x^\alpha):x\ge 0\} is contained in the closed semialgebraic set B=\{(x,y)\in\mathbb R^2:x\le 2,\; y\le 3\}\... 3 Here I record an answer based on very kind comments above (and offline). The construction starts from a closed ball B\subset\mathbb{R}^3 and an infinite sequence \mathcal{L}:=\{L^+_k\} of open half-spaces, satisfying the condition$$B\cap L^+_i\cap L^+_j=\emptyset\quad\text{ if and only if $i\neq j$.}\qquad\qquad (*)$$Denote L^-_k:=\mathbb{R}^3\... 3 Let \mathrm{Sym}^2(\mathbb{R}^n) be the \binom{n+1}{2} dimensional vector space of n \times n symmetric matrices and let P \subset \mathrm{Sym}^2(\mathbb{R}^n) be the cone of positive semidefinite matrices. A spectrahedron is the intersection of P with an affine linear space L. One reasonable interpretation of the "almost always" in your question ... 2 This is more of an extended comment than an answer, but here goes. It might be easier to consider the problem locally and then argue how global convexity follows. Local considerations don't tell if D is connected. Let f(x,y)=\Delta(x,y)/16x^2y^2. Let z=(x,y)\in\partial D and let \nu(z) be the unit outer normal at z. Take any vector a orthogonal ... 2 It is semi-algebraic. Let x be an f_0-tuple of points in R^n. Take any subset of n points from x, and find a hyperplane through these points. This involves only rational operations on the coordinates. Select from these hyperplanes those for which all points in x are on one side. This involves checking some inequalities. So we obtain equations of ... 2 The answer is yes, but not in the most exciting way. The idea is to write the envelope in the first-order logic and then kill the problem with quantifier elimination for the theory of real-closed fields. For example, we can describe the envelope as the set of points (z, w) such that for all x, y\in [a, b] we have (z, w) is above or on the segment ... 2 As Zach Teitler hinted, I shouldn't have checked it visually. In fact, if you make the construction correctly, it is quite easy to show that you have 8 real intersection with one of the ellipse and one component of the M-curve: For instance, if we use four horizontal lines all lying in the y-range of say the above arc. Then by the nature of the position of ... 1 I will jot down some stray (and easy) thoughts, in an effort to engage the question and see whether some aspects of it can be made more precise. Going out on a limb, I suppose a baseline assumption is that we start out with a collection of basic sets in \mathbb{C}^n (letting n vary) that are considered "definable", such as graphs of complex polynomial ... 1 Here I unroll the argument in [MR2399570] showing that \mathbb{L}^n=\mathbf{T}^{-1}(S) for a semialgebraic set S, to show that at worst [loc.cit.] is a bit too brief, but no more that that. Given an entrywise nonnegative n\times n matrix A, its n-tuple of eigenvalues is specified by the polynomial identity$$ \det(\lambda I -A)=\prod_k (\lambda -\...

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I cannot say much about $\geq0$. For $=0$ one can prove the following statement: if $r\leq n-1$ then then there is an $a\in \mathbf R^n\setminus\{0\}$ such that $f_i(a)=0$ for all $i$. Indeed, each polynomial $f_i$ defines a vanishing set $Z(f_i)$ in real projective $(n-1)$-space $\mathbf P^{n-1}(\mathbf R)$. Since $f_i$ is of odd degree, each vanishing set ...

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The result is known as the Pólya-Szegö theorem: Note that $\Sigma = \Sigma_n := \{ p \in \mathbb{R}[x] \mid p~\text{SOS}\}$, in which 'SOS' stands for sum of squares. A reference for the result: https://link.springer.com/chapter/10.1007/978-0-387-09686-5_7

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This seems like it can't be true since the sets $A(f_1, \ldots, f_s)$ may not be compact, which violates the precondition for outer semicontinuous, but also makes the postcondition not hold. E.g., consider $C = \{(x, y): xy-1 \ge 0\}$. It does not seem like we can force $A(f) \subseteq N_\delta(C)$ for all $f$ near $xy-1$. You will have to use some $f \in U$...

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