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Nonlinear Equations - UFRJ

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[SEC. A.3: EQUIDISTRIBUTION OF ROOTS UNDER UNITARY TRANSFORMATIONS159<br />

in a much faster way than this proved in the Bezout Series (see (A.1)<br />

below). With this new method, the Average Las Vegas algorithm<br />

was improved to have running time which is almost quadratic in the<br />

input size, see [13]. Not only the expected value of the running time<br />

is known to be polynomial in the size of the input, also the variance<br />

and other higher moments, see [16].<br />

The existence of a deterministic polynomial time algorithm for<br />

Smale’s 17th problem is still an open problem. In [25] a deterministic<br />

algorithm is shown that has running time O(N log log N ), and indeed<br />

polynomial time for certain choices of the number of variables and<br />

degree of the polynomials. There exists a conjecture open since the<br />

nineties [74]: the number of steps will be polynomial time on the<br />

average if the starting point is the homogeneization of the identity<br />

map, that is<br />

⎧<br />

z ⎪⎨<br />

d1−1<br />

0 z 1 = 0<br />

f 0 (z) = .<br />

, ζ 0 = (1, 0, . . . , 0).<br />

⎪⎩<br />

z dn−1<br />

0 z n = 0<br />

Another approach to the question is the one suggested by a conjecture<br />

in [15] on the averaging function for polynomial system solving.<br />

A.3 Equidistribution of roots under unitary<br />

transformations<br />

In the series of articles mentioned in the Smale’s 17th problem section,<br />

all the algorithms cited use linear homotopy methods for solving<br />

polynomial equations. That is, let f 1 be a (homogeneous) system to<br />

be solved and let f 0 be another (homogeneous) system which has<br />

a known (projective) root ζ 0 . Let f t be the segment from f 0 to f 1<br />

(sometimes we take the projection of the segment onto the set of systems<br />

of norm equal to 1). Then, try to (closely) follow the homotopy<br />

path, that is the path ζ t such that ζ t is a zero of f t for 0 ≤ t ≤ 1.<br />

If this path does not have a singular root, then it is well–defined. A<br />

natural question is the following: Fix f 1 and consider the orbit of f 0<br />

under the action f 0 ↦→ f 0 ◦ U ∗ where U is a unitary matrix. The root

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