如何在Python中进行非线性复杂的根查找 [英] How to do nonlinear complex root finding in Python

问题描述

我想对以下非线性方程式进行根搜索，我在Python中进行了搜索，但没有用.我的代码在下面

from pylab import *
import scipy
import scipy.optimize

def z1(x,y):
    temp=1+1j+x+2*y;
    return temp

def z2(x,y):
    temp=-1j-2*x+sqrt(3)*y;
    return temp

def func(x):
    temp=[z1(x[0],x[1])-1.0/(1-1.0/(z2(x[0],x[1]))),1-2.0/(z2(x[0],x[1])-4.0/z1(x[0],x[1]))]
    return temp

result=scipy.optimize.fsolve(func,[1+1j,1+1j])

print result

当我运行它时，它显示错误:

---> 30 result = scipy.optimize.fsolve(func，[1 + 1j，1 + 1j])

fsolve中的C:\ Python27 \ lib \ site-packages \ scipy \ optimize \ minpack.py(func，x0，args，fprime，full_output，col_deriv，xtol，maxfev，band，epsfcn，factor，diag)

123             maxfev = 200*(n + 1)

124         retval = _minpack._hybrd(func, x0, args, full_output, xtol,

-> 125 maxfev，ml，mu，epsfcn，因子，diag)

126     else:

127         _check_func('fsolve', 'fprime', Dfun, x0, args, n, (n,n))

解决方案

fsolve从R ^ n-> R中找到函数的零.类似的函数 解决方案

fsolve finds zeros of functions from R^n -> R. The similar function root finds zeros of functions from R^n -> R^m.

It looks like you're trying to find zeros of a function from C^2 -> C^2, which as far as I know scipy.optimize doesn't support directly - but you could try writing it a function from R^4 -> R^4 and then using root. For example, something along the lines of:

def func_as_reals(x):
    r1, c1, r2, c2 = x
    a, b = func([complex(r1, c1), complex(r2, c2)])
    return [a.real, a.imag, b.real, b.imag]

should work, though it might be significantly faster to do it directly on the real numbers instead of repeatedly wrapping into complex and unwrapping.

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