By Saugata Basu, Richard Pollack, Marie-Francoise Roy,
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Additional resources for Algorithms in Real Algebraic Geometry, Second Edition (Algorithms and Computation in Mathematics)
E. of the form 2m−1n with n odd. By induction hypothesis, it has a root γ in R[i]. Since the classical method for solving polynomials of degree 2 works in R[i] when R is real closed, the roots of the second degree polynomial F (z, γ)X 2 − G(z, γ)X + H(z, γ) are roots of P that belong to R[i]. We have proved that the polynomial P has a root in R[i]. For P = a p X p + + a0 ∈ R[i][X], we write P = a p X p + + a0. Since P P ∈ R[X], P P has a root x in R[i]. Thus P (x) = 0 or P (x) = 0. In the ﬁrst case we are done and in the second, P (x) = 0.
36. 35 (Descartes’s law of signs). 37. The polynomial P = X 2 − X + 1 has no real root, but Var(Der(P ); 0, 1) = 2. It is impossible to ﬁnd a ∈ (0, 1] such that Var(Der(P ); 0, a) = 1 and Var(Der(P ); a, 1) = 1 since otherwise P would have two real roots. This means that however we reﬁne the interval (0, 1], we are going to have an interval (the interval (a, b] containing 1/2) giving 2 sign variations. 12. Prove that − If Var(Der(P ); a, b) = 0, then P has no root in (a, b]. − If Var(Der(P ); a, b) = 1, then P has exactly one root in (a, b], which is simple.
Proof: Suppose that P = (X − x) µ Q and Q(x) 0. It is clear that P (x) = 0. The proof of the claim is by induction on the degree of P . The claim is obviously true for deg(P ) = 1. Suppose that the claim is true for every polynomial of degree < d. Since P = (X − x) µ−1 (µ Q + (X − x) Q ), and µ Q(x) 0, by induction hypothesis, P (x) = = P (µ−1)(x) = 0, P (µ)(x) 0. Conversely suppose that P (x) = P (x) = = P (µ−1)(x) = 0, P (µ)(x) 0. 1 (Taylor’s formula) at x, P = (X − x) µ Q, with Q(x) = P (µ)(x)/µ!
Algorithms in Real Algebraic Geometry, Second Edition (Algorithms and Computation in Mathematics) by Saugata Basu, Richard Pollack, Marie-Francoise Roy,