User michel coste - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:53:47Z http://mathoverflow.net/feeds/user/2630 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9431/when-does-a-real-polynomial-have-a-pair-of-complex-conjugate-roots/9444#9444 Answer by Michel Coste for When does a real polynomial have a pair of complex conjugate roots? Michel Coste 2009-12-20T21:53:59Z 2009-12-20T22:43:41Z <p>We can assume $f$ has no multiple root (if the gcd of $f$ and $f'$ is not constant, divide by this gcd). Let $n$ be the degree of $f$. Compute $$\frac{f(X)f'(Y)-f(Y)f'(X)}{X-Y} = \sum_{i,j=i}^{n}a_{i,j}\; X^{i-1}\; Y^{j-1}\;.$$ Then $f$ has all roots real iff the symmetric matrix $(a_{i,j})_{i,j=1,\ldots,n}$ is positive definite. This can be checked for instance by computing the principal minors of this matrix and verifying whether they are all positive.</p> <p>There are several methods for computing the number of real roots using signature of quadratic form : see for instance <a href="http://ens.univ-rennes1.fr/agreg-maths/documentation/docs/racsign.pdf" rel="nofollow">this note</a> (in french).</p> http://mathoverflow.net/questions/9356/real-spectrum-of-ring-of-continuous-semialgebraic-functions/9363#9363 Answer by Michel Coste for Real spectrum of ring of continuous semialgebraic functions Michel Coste 2009-12-19T09:44:48Z 2009-12-19T09:44:48Z <p>I don't agree with the preceding answer.</p> <p>When $U$ is a locally compact semialgebraic set, then $\widetilde{U}$ equipped with its sheaf of semi-algebraic continuous functions is isomorphic to the affine scheme $\mathrm{Spec}(S^0(U))$. This is proposition 6 in Carral, Coste : Normal spectral spaces and their dimensions, J. Pure Appl. Algebra 30 (1983) 227-235. In particular $\widetilde{U}$ is homeomorphic to the prime spectrum of $S^0(U)$, which is homeomorphic to its real spectrum. In case $U$ is not locally compact, the situation is different; there are more points in $\mathrm{Spec}(S^0(U))$.</p> http://mathoverflow.net/questions/9089/what-was-hilberts-geometric-construction-in-his-17th-problem/9124#9124 Answer by Michel Coste for what was Hilbert's geometric construction in his 17th problem? Michel Coste 2009-12-16T19:34:11Z 2009-12-16T19:34:11Z <p>Actually the answer is in the sections 36 to 39 of Hilbert's "Foundations of geometry", which can be found on the web. The constructions are construction with "straightedge" (ruler) and "transferrer of segments". I quote a result from Hilbert's book :</p> <p>Theorem 41. A problem in geometrical construction is, then, possible of solution by the drawing of straight lines and the laying off of segments, that is to say, by the use of the straight-edge and a transferrer of segments, when and only when, by the analytical solution of the problem, the co-ordinates of the desired points are such functions of the co-ordinates of the given points as may be determined by the rational operations and, in addition, the extraction of the square root of the sum of two squares.</p> <p>This result explains relatively clearly why this kind of geometrical constructions leads to the question of the determination of those functions of $x_1,\ldots,x_n$ which can be written as sums of squares of rational functions with rational coefficients.</p> http://mathoverflow.net/questions/9356/real-spectrum-of-ring-of-continuous-semialgebraic-functions/9363#9363 Comment by Michel Coste Michel Coste 2009-12-19T14:08:28Z 2009-12-19T14:08:28Z I am not quite sure of what you call Harrison topology. The topology on $\widetilde U$ has for basis the $\widetilde V$ for $V$ semialgebraic open in $U$. The topology on the real spectrum of $S^0(U)$ has for (sub)basis the $\{\alpha\mid f(\alpha)&gt;0\}$. These are different from the constructible topology.