[math-fun] Universal point-set for polynomial rootfinding
http://mathlab.mathlab.sunysb.edu/~scott/Papers/Newton/Published.pdf This paper is quite interesting. Given any polynomial P(z) of degree=D such that we know all its roots lie within the unit circle, they describe a "universal" set of 2.22 * D * (lnD)^2 points such that Newton's method starting from those points is guaranteed to converge to all the roots of P. By "universal" I mean that this point set does not know or care what P is (aside from knowing its degree and knowing al its roots lie in the unit disk). The point set is: Let N=ceiling(8.32547*D*lnD) and M=ceiling(0.26632*lnD). Equispace N points around the perimeters of each of the following M circles centered at z=0: radius = (1+sqrt(2)) * ((D-1)/D)^((2*v-1)/(4*M)) for each v=1,2,3,...,M. That's all.
This does look very nice --- I think Henry Baker mentioned the graphics, but more than just pretty, they show very clearly how the idea works. It always impresses me when somebody finds something (apparently) fresh to say about an apparently bog-standard, thoroughly familiar problem. Henrici (1974) proposed a simultaneous rootfinder based on circle arithmetic --- I wonder how the perfomance of that compares with this in practice? WFL On 2/16/12, Warren Smith <warren.wds@gmail.com> wrote:
http://mathlab.mathlab.sunysb.edu/~scott/Papers/Newton/Published.pdf
This paper is quite interesting. Given any polynomial P(z) of degree=D such that we know all its roots lie within the unit circle, they describe a "universal" set of 2.22 * D * (lnD)^2 points such that Newton's method starting from those points is guaranteed to converge to all the roots of P.
By "universal" I mean that this point set does not know or care what P is (aside from knowing its degree and knowing al its roots lie in the unit disk).
The point set is: Let N=ceiling(8.32547*D*lnD) and M=ceiling(0.26632*lnD). Equispace N points around the perimeters of each of the following M circles centered at z=0: radius = (1+sqrt(2)) * ((D-1)/D)^((2*v-1)/(4*M)) for each v=1,2,3,...,M.
That's all.
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participants (2)
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Fred lunnon -
Warren Smith