Python:寻找非线性方程的多个根 [英] Python: Finding multiple roots of nonlinear equation

问题描述

假设函数如下:

f(x) = x * cos(x-4)

With x = [-2.5, 2.5] 这个函数在 f(0) = 0 和 f(-0.71238898) = 0.

这是通过以下代码确定的:

导入数学从 scipy.optimize 导入 fsolve定义函数(x):返回 x*math.cos(x-4)x0 = fsolve(func, 0.0)# 返回 [0.]x0 = fsolve(func, -0.75)# 返回 [-0.71238898]

使用 fzero(或任何其他 Python 根查找器)在一次调用中查找两个根的正确方法是什么?是否有不同的 scipy 函数可以做到这一点?

fzero 参考

解决方案

定义您的函数，使其可以接受标量或 numpy 数组作为参数:

<预><代码>>>>将 numpy 导入为 np>>>f = lambda x : x * np.cos(x-4)

然后将参数向量传递给 fsolve.

<预><代码>>>>x = np.array([0.0, -0.75])>>>fsolve(f,x)数组([ 0., -0.71238898])

Assume the following function:

f(x) = x * cos(x-4)

With x = [-2.5, 2.5] this function crosses 0 at f(0) = 0 and f(-0.71238898) = 0.

This was determined with the following code:

import math
from scipy.optimize import fsolve
def func(x):
    return x*math.cos(x-4)
x0 = fsolve(func, 0.0)
# returns [0.]
x0 = fsolve(func, -0.75)
# returns [-0.71238898]

What is the proper way to use fzero (or any other Python root finder) to find both roots in one call? Is there a different scipy function that does this?

fzero reference

解决方案

Define your function so that it can take either a scalar or a numpy array as an argument:

>>> import numpy as np
>>> f = lambda x : x * np.cos(x-4)

Then pass a vector of arguments to fsolve.

>>> x = np.array([0.0, -0.75])
>>> fsolve(f,x)
array([ 0.        , -0.71238898])

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