By Morris Marden
Through the years because the first version of this famous monograph seemed, the topic (the geometry of the zeros of a posh polynomial) has endured to exhibit an analogous awesome energy because it did within the first a hundred and fifty years of its heritage, starting with the contributions of Cauchy and Gauss. hence, the variety of entries within the bibliography of this variation needed to be elevated from approximately three hundred to approximately six hundred and the booklet enlarged by way of one 3rd. It now features a extra broad remedy of Hurwitz polynomials and different issues. the recent fabric on infrapolynomials, summary polynomials, and matrix equipment is of specific curiosity.
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Extra resources for Geometry of Polynomials
46 INVARIANTIVE FORMULATION  FIG. (11,1) By drawing lines from the north pole N of S through the points P and P; let us project P and P; stereographically upon the equatorial plane of S into the points z and z, , respectively. At the points z, let us place masses m, which repel (attract if m, < 0) a unit mass at z according to the inverse distance law. Let us denote by O(z) the resultant force at z. We ask: what is the relation between the resultant force (D(P) in the spherical field and the resultant force O(z) in the corresponding plane field ?
EXERCISES. Prove the following. 1. If the zeros of a polynomial f (z) are symmetric in a line L, then between two successive zeros of f(z) on L lie an odd number of zeros of its derivative f(z) and any interval of L which contains all the zeros of f(z) lying on L also contains all the zeros of f'(z) lying on L. Hint: Apply (10,7) to Rolle's Theorem. 2. Let z = g(Z) be a rational function which has as its only poles those of , k. Let furthermore multiplicities qr at. the points Q; with j = 1, 2, h(Z) = H7=1 (Z - Q)°; and F(Z) = h(Z)'nJ'(g(Z)), where f(z) is the function (10,1).
INFRAPOLYNOMIALS 17 Writing p(z) = z" + a1z"-1 + + a,,, multiplying the kth equation in (5,10) by ak for each k and adding the resulting equations, we are led to the further equation [P(Zj) - 41 = 0 MAj P(zj) j=o which because I Aj = 1 is the same as (5,11) DAZ 3I P(Zj)] = 1. J=O The n + 1 equations (5,11) and (5,10) may be regarded as involving the m + 1 unknowns Aj of which at least one is different from zero. k = V(ZO, Z1, ... I Zk) P(ZO)P(Z1) ... P(Z") where V(zo , z1 , , zk), as the Vandermonde determinant for the distinct numbers zo , z1 , , zk , is different from zero.