一道对数题

无心人

确实可以，但牛顿迭代如果带入复数，存在进入某个循环圈的可能

winxos

原帖由 无心人 于 2009-4-2 08:45 发表 另外，好像有另外一个定理能定界根的范围 在我一本数学小辞典上 谁清楚，在这里贴出来 完善下这个帖子 呵呵 ==================================== 是否有必要拆帖子 好像偏离题目了

我以前也在哪里见过解高次方程的，好像是一本计算机方面的书，也提到了解的范围问题，想不起来哪里了，晕。

wayne

给大家看看牛顿分形（Newton fractal） http://en.wikipedia.org/wiki/Newton_fractal

The Newton fractal is a boundary set in the complex plane which is characterized by Newton's method applied to a fixed polynomial . It is the Julia set of the meromorphic function [img=http://upload.wikimedia.org/math/9/6/b/96b12b98eeba57314f362b1420ac491b.png]z\mapsto z-\tfrac{p(z)}{p'(z)}[/img] which is given by Newton's method. When there are no attractive cycles, it divides the complex plane into regions Gk, each of which is associated with a root ζk of the polynomial, [img=http://upload.wikimedia.org/math/7/e/0/7e039dc273650c591d637ad2f4176bdb.png]k=1,..,\operatorname{deg}(p)[/img]. In this way the Newton fractal is similar to the Mandelbrot set, and like other fractals it exhibits a complex appearance arising from a simple description. It is relevant to numerical analysis because it shows that (outside the region of quadratic convergence) the Newton method can be very sensitive to its choice of start point. Many points of the complex plane are associated with one of the [img=http://upload.wikimedia.org/math/0/4/1/041165f3457932e25c874d3d5dee0f4c.png]\operatorname{deg}(p)[/img] roots of the polynomial in the following way: the point is used as starting value z0 for Newton's iteration [img=http://upload.wikimedia.org/math/7/3/c/73c2e10f2885e66f9b5b947cdf72bba6.png]z_{n+1}:=z_n-\frac{p(z_n)}{p'(z_n)}[/img], yielding a sequence of points z1, z2, .... If the sequence converges to the root ζk, then z0 was an element of the region Gk. However, for every polynomial of degree at least 2 there are points for which the Newton iteration does not converge to any root: examples are the boundaries of the basins of attraction of the various roots. There are even polynomials for which open sets of starting points fail to converge to any root: a simple example is z3 − 2z + 2, where some points are attracted by the cycle 0, 1, 0, 1 ... rather than by a root. To plot interesting pictures, one may first choose a specified number d of complex points (ζ1,...,ζd) and compute the coefficients (p1,...,pd) of the polynomial [img=http://upload.wikimedia.org/math/3/0/1/3014425872ca80f3e238ff3e9a37f388.png]p(z)=z^d+p_1z^{d-1}+\cdots+p_{d-1}z+p_d:=(z-\zeta_1)\cdot\cdots\cdot(z-\zeta_d)[/img]. Then for a rectangular lattice zmn = z00 + mΔx + inΔy, m = 0, ..., M − 1, n = 0, ..., N − 1 of points in [img=http://upload.wikimedia.org/math/f/0/b/f0b01fe0a1eec87c634584ac0694fb71.png]\mathbb{C}[/img], one finds the index k(m,n) of the corresponding root ζk(m,n) and uses this to fill an M×N raster grid by assigning to each point (m,n) a colour fk(m,n). Additionally or alternatively the colours may be dependent on the distance D(m,n), which is defined to be the first value D such that | zD − ζk(m,n) | < ε for some previously fixed small ε > 0. Generalization of Newton fractals A generalization of Newton's iteration is [img=http://upload.wikimedia.org/math/9/0/9/9094eb2ebbb177cb280fe221a653e6fe.png]z_{n+1}=z_n- a \frac{p(z_n)}{p'(z_n)}[/img] where a is any complex number[1]. The special choice a = 1 corresponds to the Newton fractal. The fixed points of this map are stable when a lies inside the disk of radius 1 centered at 1. When a is outside this disk, the fixed points are locally unstable, however the map still exhibits a fractal structure in the sense of Julia set.If p is a polynomial of degree n, then the sequence zn is bounded provided that a is inside a disk of radius n centered at n. More generally, Newton's fractal is a special case of a Julia set

Newton fractal for three degree-3 roots (p(z) = z^3 − 1), coloured by number of iterations required	Newton fractal for three degree-3 roots (p(z) = z^3 − 1), coloured by root reached	Newton fractal for p(z) = z^3 − 2z + 2. Points in the red basins do not reach a root.	Newton fractal for x^8 + 1 − 16
Newton fractal for p(z) = z^5 − 3i z^3 − (5 + 2i)z^2 + 3z + 1, coloured by root reached, shaded by number of iterations required	Newton fractal for p(z) = sin(z), coloured by root reached, shaded by number of iterations required	Another Newton fractal for sin(x)	Generalized Newton fractal for p(z) = z^3 − 1, a = − 0.5. The colour was chosen based on the argument after 40 iterations.
Generalized Newton fractal for p(z) = z^2 − 1, a = 1 + i.	Generalized Newton fractal for p(z) = z^3 − 1, a = 2.	Generalized Newton fractal for p(z) = z^4 + 3i − 1, a = 2.1.

[ 本帖最后由 wayne 于 2009-4-23 12:17 编辑 ]