複素係数代数方程式の数値解法
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概要
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On the complex plane we examine if a given polynomial f(z) has all roots inside a given circle of radius R. If it has not, then we double the radius and examine similarly. Soon we find the circle containing all roots of f(z). Assuming that there are all roots of f(z) inside the circle of radius 2R, the largest root of an absolute value exists at least in an annulus R<∣z∣<2R. Here we put R_<in>=R and R_<ex>=2R. We examine if f(z) has all roots inside a circle of radius R_m which is equal to (R_<in>+R_<ex>)/2. If it has, then we narrow an annulus to R_<in><∣z∣<R_m, if not, then to R_m<∣z∣<R_<ex>. Continuing this procedure, finally we are able to lay a root on the circle of radius R_m such that R_<in>=R_m=R_<ex>. Then R_m is equal to an absolute value of the root. Next, for finding a place where the root on the circle lies, we transform z to R_mz, then the circle of radius R_m is replaced by the unit circle Γwhose radius is equal to one, construct a sequence of Lehmer's polynomial so that f_0(R_mz)=f(R_mz), f_1(R_mz), …, f_h(R_mz), f_<h+1> where f_<h+1>=0,f_h(R_mz) has all roots on Γ. Here we put p(z)=f_h(R_mz), and consider finding all roots of p(z) on Γ. Now we move the center of coordinate to a point ce^<iθ> outside Γ so that z is replaced by z'+ce_<iθ>. As before we examine if p(z') has all roots inside a circle of radius r. If it has not, then we double r. Soon we find an annulus r<∣z'∣<2r which contains at least the largest root of an absolute value. Here we put r_<in>=r and r_<ex>=2r. We examine if p(z') has all roots inside a circle of radius r_m which is equal to (r_<in>+r_<ex>)/2. If it has, then we narrow an annulus to r_<in><∣z'∣<r_m, if not, then to r_m<∣z'∣<r_<ex>. Continuing as far as possible this procedure, finally a root lies on the circle Γ' of radius r_m such that r_<in>=r_m=r_<ex>. Then Γ and Γ' cross each other. Thus a root exists in either of two crossing places so that z_1' and z_2'. If ∣p(z_1')∣<∣p(z_2')∣, then z_1' is a root of p(z'), if, instead, ∣p(z_1')∣>∣p(z_2')∣, then z_2' is so. Hence we find a root of f(z) so that z=(z'+ce^<iθ>R_m.