找出具有约束的方程式的根 [英] Finding the root of an equation with a constraint

问题描述

在python中，我想找到以下形式的方程的根:

In python, I would like to find the roots of equations of the form:

-x * log(x)+(1-x)* log(n)-(1-x)* log(1-x)-k = 0

-x*log(x) + (1-x)*log(n) - (1-x)*log(1 - x) - k = 0

其中n和k是将要指定的参数.

where n and k are parameters that will be specified.

对根的另一个约束是x> =(1-x)/n.因此，就其价值而言，我将过滤掉不满足要求的根.

An additional constraint on the roots is that x >= (1-x)/n. So just for what it's worth, I'll be filtering out roots that don't satisfy that.

我的第一个尝试是使用scipy.optimize.fsolve(请注意，我只是将k和n分别设置为0和1):

My first attempt was to use scipy.optimize.fsolve (note that I'm just setting k and n to be 0 and 1 respectively):

def f(x):                                                                       
    return -x*log(x) + (1-x)*log(1) - (1-x)*log(1-x)                            

fsolve(f, 1)

使用math.log时，我遇到了价值错误，因为我向日志提供了错误的输入.使用numpy.log给了我一些零除和乘法中的无效值.

Using math.log, I got value-errors because I was supplying bad input to log. Using numpy.log gave me some divide by zeros and invalid values in multiply.

我这样调整了f，只是看它会做什么:

I adjusted f as so, just to see what it would do:

def f(x):                                                                       
    if x <= 0:                                                                  
        return 1000                                                             
    if x >= 1:                                                                  
        return 2000                                                             
    return -x*log(x) + (1-x)*log(1) - (1-x)*log(1-x) 

现在我得到

/usr/lib/python2.7/dist-packages/scipy/optimize/minpack.py:221: RuntimeWarning: The iteration is not making good progress, as measured by the 
  improvement from the last ten iterations.
  warnings.warn(msg, RuntimeWarning)     

使用python，如何求解原始方程式中各种n和k参数的x?

Using python, how can I solve for x for various n and k parameters in the original equation?

推荐答案

fsolve还允许插入从何处开始的猜测.我的建议是绘制方程式，并让用户使用鼠标或通过文本输入初始猜测值作为初始猜测值.您可能还需要更改超出范围的值:

fsolve also allows guesses to be inserted for where to start. My suggestion would be to plot the equation and have the user type a initial guess either with the mouse or via text to use as an initial guess. You may also want to change the out of bounds values:

if x <= 0:
    return 1000 + abs(x)
if x >= 1:
    return 2000 + abs(x)

通过这种方式，该函数在感兴趣区域之外具有一个斜率，它将引导求解器返回到感兴趣的区域.

This way the function has a slope outside of the region of interest that will guide the solver back into the interesting region.

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