用python求解超越方程组 [英] Solving a system of transcendental equations with python

问题描述

假设我有以下四个等式:

Assuming I have the following four equations:

1. cos(x)/x = a
2. cos(y)/y = b
3. a + b = 1
4. c sinc(x) = d sinc(y)

对于未知变量x, y, a 和b.请注意，cos(x)/x=a 有多种解决方案.变量 y 也类似.我只对 x 和 y 值感兴趣，它们是第一个正根(如果重要的话).

for unknown variables x, y, a and b. Note that cos(x)/x=a has multiple solutions. Similar goes for variable y. I am only interested in x and y values, which are first positive roots (if that matters).

您可以安全地假设 a, b, c 和 d 是已知的实常数，都是正的.

You can safely assume a, b, c and d are known real constants, all positive.

在 Mathematica 中，解决这个问题的代码如下所示:

In Mathematica the code to solve this would look something like:

FindRoot[{Cos[x]/x == 0.2 a + 0.1, 
          Cos[y]/y == 0.2 b + 0.1, 
          a + b == 1.0, 
           1.03*Sinc[x] == Sinc[y]*1.02}, 
          {{x, .1}, {y, .1}, {a, .3}, {b, .1}}]

结果返回

{x -> 1.31636, y -> 1.29664, a -> 0.456034, b -> 0.543966}

虽然这很容易，但我不知道如何在 python 中做这样的事情.因此，如果有人能指导我(或简单地告诉我如何)解决这个问题，我将不胜感激.

While this was quite easy, I have no idea how to do anything like that in python. So if somebody could kinda guide me (or simply show me how) to solve this, I would highly appreciate it.

推荐答案

您可以使用 root:

You can use root:

import numpy as np
from scipy.optimize import root

def your_funcs(X):

    x, y, a, b = X

    f = [np.cos(x) / x - 0.2 * a - 0.1,
         np.cos(y) / y - 0.2 * b - 0.1,
         a + b - 1,
         1.03 * np.sinc(x) - 1.02 * np.sinc(y)]

    return f

sol2 = root(your_funcs, [0.1, 0.1, 0.3, 0.1])
print(sol2.x)

会打印

[ 1.30301572  1.30987969  0.51530547  0.48469453]

您的函数必须以求值为 0 的方式定义，例如a + b - 1 而不是 a + b = 1.

Your functions have to be defined in a way that they evaluate to 0, e.g. a + b - 1 instead of a + b = 1.

快速检查:

print(your_funcs(sol2.x))

给予

[-1.9356960478944529e-11, 1.8931356482454476e-11, 0.0, -4.1039033282785908e-11]

所以，解应该没问题(请注意e-11基本为0).

So, the solution should be ok (please note that e-11 is basically 0).

或者，您也可以使用 fsolve:

Alternatively, you can also use fsolve:

from scipy.optimize import fsolve

sol3 = fsolve(your_funcs, [0.1, 0.1, 0.3, 0.1])

给你同样的结果:

[ 1.30301572  1.30987969  0.51530547  0.48469453]

您可以使用 args 参数传递其他参数:

You can pass additional arguments using the args argument:

def your_funcs(X, fac_a, fac_b):

    x, y, a, b = X

    f = [np.cos(x) / x - fac_a * a - 0.1,
         np.cos(y) / y - fac_b * b - 0.1,
         a + b - 1,
         1.03 * np.sinc(x) - 1.02 * np.sinc(y)]

    return f

sol2 = root(your_funcs, [0.1, 0.1, 0.3, 0.1], args=(0.2, 0.2))
print(sol2.x)

它为您提供旧"输出:

[ 1.30301572  1.30987969  0.51530547  0.48469453]

如果你跑

sol2 = root(your_funcs, [0.1, 0.1, 0.3, 0.1], args=(0.4, 0.2))
print(sol2.x)

然后你会收到:

[ 1.26670224  1.27158794  0.34096159  0.65903841]

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