我正试图解一组简单的方程。手动找到解决方案很简单,但是我想用同情来学习这个工具。
from sympy import symbols,solve,Le,Eq
l,x = symbols('lamda x')
f0 = x**2+1
f1 = (x-2)*(x-4); feasible_set = Le(f1,0);
lagrange = f0 + l*f1
stationary_lagrangian = Eq(lagrange.diff(x),0)
solve([feasible_set,stationary_lagrangian])上面的代码给出了错误NotImplementedError: inequality has more than one symbol of interest.。
问题1:为什么会这样?不等式只包含x,而不包含lamda。
问题2:是否有可能用另一种方式解决同样的问题?
问题的背景,如果你有兴趣的话
minimize (over x \in R)
x^2 + 1
subject to
(x-2)(x-4) <= 0。。然后从KKT条件中应用平稳性和原始可行性。
发布于 2019-04-12 12:43:22
正如注释中提到的那样,sympy.solve解决了等式系统的问题。所以应该是,
from sympy import solve, var, symbols, diff
x = var('x',real=True);
f = x**2+1
g = (x-2)*(x-4)
l = symbols('lambda', real = True)
lagrange = f - l* g
grad = [diff(lagrange,x)]
kkt_eqs = grad + [g]
extremum_points = solve(kkt_eqs, [x, l], dict=True) 编辑:现在,从极值点,你必须找到最小。
f_x_ = min(ele[x]**2 + 1 for ele in stationary_points)
minimum = [ele[x] for ele in stationary_points if ele[x]**2+1 == f_x_]
print(minimum)https://stackoverflow.com/questions/55648586
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