Posted in Python onMay 07, 2015
本文实例讲述了python计算方程式根的方法。分享给大家供大家参考。具体实现方法如下:
''' roots = polyRoots(a). Uses Laguerre's method to compute all the roots of a[0] + a[1]*x + a[2]*x^2 +...+ a[n]*x^n = 0. The roots are returned in the array 'roots', ''' from evalPoly import * from numpy import zeros,complex from cmath import sqrt from random import random def polyRoots(a,tol=1.0e-12): def laguerre(a,tol): x = random() # Starting value (random number) n = len(a) - 1 for i in range(30): p,dp,ddp = evalPoly(a,x) if abs(p) < tol: return x g = dp/p h = g*g - ddp/p f = sqrt((n - 1)*(n*h - g*g)) if abs(g + f) > abs(g - f): dx = n/(g + f) else: dx = n/(g - f) x = x - dx if abs(dx) < tol: return x print 'Too many iterations' def deflPoly(a,root): # Deflates a polynomial n = len(a)-1 b = [(0.0 + 0.0j)]*n b[n-1] = a[n] for i in range(n-2,-1,-1): b[i] = a[i+1] + root*b[i+1] return b n = len(a) - 1 roots = zeros((n),dtype=complex) for i in range(n): x = laguerre(a,tol) if abs(x.imag) < tol: x = x.real roots[i] = x a = deflPoly(a,x) return roots raw_input("\nPress return to exit")
希望本文所述对大家的Python程序设计有所帮助。
python计算方程式根的方法
- Author -
songguo声明:登载此文出于传递更多信息之目的,并不意味着赞同其观点或证实其描述。
Reply on: @reply_date@
@reply_contents@