By Gonzalo A. Aranda-Corral, Jacques Calmet, Francisco J. Martín-Mateos
This publication constitutes the refereed complaints of the twelfth foreign convention on man made Intelligence and Symbolic Computation, AISC 2014, held in Seville, Spain, in December 2014. The 15 complete papers awarded including 2 invited papers have been conscientiously reviewed and chosen from 22 submissions. The objectives have been on one facet to bind mathematical domain names akin to algebraic topology or algebraic geometry to AI but additionally to hyperlink AI to domain names outdoors natural algorithmic computing. The papers deal with all present facets within the region of symbolic computing and AI: simple techniques of computability and new Turing machines; logics together with non-classical ones; reasoning; studying; determination help platforms; and desktop intelligence and epistemology and philosophy of symbolic mathematical computing.
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Additional resources for Artificial Intelligence and Symbolic Computation: 12th International Conference, AISC 2014, Seville, Spain, December 11-13, 2014. Proceedings
Xn ) = P (X12 , . . , Xn2 ) is Psd (Rn ) (note that this transformation doubles the degree of P ), we can use the following result. Proposition 2 (Hilbert).  If P ∈ R[X] is Psd (R), then P is a sum of two squares of polynomials. Example 2. Consider H(x) = Q(x2 ) = x6 − 4x4 + 6x2 + 1 = f1 (x) + f2 (x) where fi (x) = (ai x3 + bi x2 + ci x + di )2 for i = 1, 2. , Mathematica. We have V (n) = 2n + 2 and I(n) = 2n + 1. 2 P´ olya and Szeg¨ o Proposition 3 (P´ olya & Szeg¨ o).  If P is Psd ([0, +∞)), then there are sos polynomials f, g such that P (x) = f (x) + xg(x) and deg(f ), deg(xg) ≤ deg(P ).
We have V (n) = n and I(n) = n + 1 + V (n) = 2n + 1. 4 Hilbert with Gram Matrices An alternative way to use Hilbert’s representation is the following. Theorem 2.  Let P be a polynomial of degree 2m and z(X) be the vector of all monomials X α such that |α| ≤ m. Then, P is a sum of squares in R[X] if and only if there exists a real, symmetric, psd matrix B such that P = z(X)T Bz(X). Proving H(x) = P (x2 ) of degree 2n to be sos amounts at (1) matching H against z(X)T Bz(X) (where z(X) = (1, X, .
Then, we need to solve 2n + 1 equations in variables 2 Using Representation Theorems for Proving Polynomials Non-negative 25 (the parameters bij ) corresponding to the monomials in H. , B must be positive semideﬁnite. This can be done by computing the characteristic polynon mial det (zIn+1 − B) = i=0 ci z i of B and requiring its roots to be non-negative . They show that this can be achieved by imposing (−1)i+n+1 ci ≥ 0 for all 0 ≤ i ≤ n. Thus, V (n) = (n+1)(n+2) . and I(n) = (2n + 1) + (n + 1) = 3n + 2.