有人告诉我,只要提供一阶导数,方法scipy.optimize.newton()就能解决复杂的功能.我不能使它工作. newton()的文档中没有提到任何复杂的函数.有人可以告诉我如何在SciPy中找到f(z)= 1 z ^ 2之类的函数的根吗？我需要解决更复杂的问题,但是一个简单的示例将极大地帮助我.

解决方法:

这是在IPython会话中将newton与复杂函数结合使用的示例：

In [1]: def func(z):
   ...:     return 1 + z*z
   ...: 

In [2]: def deriv(z):
   ...:     return 2*z
   ...: 

In [3]: from scipy.optimize import newton

In [4]: newton(func, x0=1+1j, fprime=deriv, tol=1e-12)
Out[4]: 1j

In [5]: newton(func, x0=-2j, fprime=deriv, tol=1e-12)
Out[5]: -1j

请注意,牛顿不将x0处理为导数为零的点：

In [6]: newton(func, x0=0, fprime=deriv, tol=1e-12)
/Users/warren/local_scipy/lib/python2.7/site-packages/scipy/optimize/zeros.py:119: RuntimeWarning: derivative was zero.
  warnings.warn(msg, RuntimeWarning)
Out[6]: 0.0

另外,如果您的函数返回实参的实数值,请确保为x0传递一个复数值.否则,该函数将卡在实轴上：

In [21]: newton(func, x0=0.5, fprime=deriv, tol=1e-12)
---------------------------------------------------------------------------
RuntimeError                              Traceback (most recent call last)
<ipython-input-21-2feb08057c57> in <module>()
----> 1 newton(func, x0=0.5, fprime=deriv, tol=1e-12)

/Users/warren/local_scipy/lib/python2.7/site-packages/scipy/optimize/zeros.pyc in newton(func, x0, fprime, args, tol, maxiter, fprime2)
    159             q1 = func(*((p1,) + args))
    160     msg = "Failed to converge after %d iterations, value is %s" % (maxiter, p)
--> 161     raise RuntimeError(msg)
    162 
    163 

RuntimeError: Failed to converge after 50 iterations, value is -0.870752774435

增加maxiter将无济于事：

In [22]: newton(func, x0=0.5, fprime=deriv, tol=1e-12, maxiter=1000)
---------------------------------------------------------------------------
RuntimeError                              Traceback (most recent call last)
[...]
RuntimeError: Failed to converge after 1000 iterations, value is -0.0895687261655