Re: Homotopic algorithm to solve a system of equations

*To*: mathgroup at smc.vnet.net*Subject*: [mg66875] Re: Homotopic algorithm to solve a system of equations*From*: "aTn" <ayottes at dms.umontreal.ca>*Date*: Fri, 2 Jun 2006 04:09:01 -0400 (EDT)*References*: <e5mhml$kjl$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Ok, some of you wrote back to me wondering if such algorithms exist and how they work. Here is a quick explanation. The methods I'm referring to are know as "probability one globally convergent homotopy algorithms" (see L. Waston's great introductory paper) that were developped to help solve a system of equations (polynomial with real coefficients). The basic result used in these algorithms is a parametrized version of Sards theorem: Let U be an open set in R^m and let p: U x R^n x [0,1[ ---> R^n be a C^2 map. If p is transversal to zero (i.e. 0 is a regular value of p), then the set of points 'v' in the open set U such that p_v : R^n x [0,1[ --> R^n : (x,t) ---> p(v,x,t) is transversal to 0, is of measure one. The way to apply this result to equation solving is the following. Consider a finite set of n polynomials with n indeterminates (and real coefficients). To it corresponds a (smooth) function F:R^n---> R^n. The idea is to find an open set U in R^m, a C^2 map p: U x R^n x [0,1[ --> R^n which is transversal to zero, such that a solution of p_v(x,0) = 0 is known explicitly for all v in U, p_v(x,1)= F(x) for all v in U and all x in R^n, and finally for which p_v^(-1)(0) is bounded for all v in U. Then (by the previous theorem) you can find a path in p_a^(-1)(0) for some 'v' for which the differential of p_v is of full rank linking a zero of p_v(. , 0) to a zero of p_v(. , 1) = F. We want the differential to be of full rank so we can use algorithms such as the Newton method to solve successive problems along the chosen path. I'm looking for a package implementing known probability one globally convergent homotopy algorithms. I hope this message makes my question clearer. Regards, Etienne (aTn) P.S.: I typed fast, so feel free to point out mystakes I might have made :)

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