Question

我有一个多项式，我试图找到所有根以使用nsolve进行数值计算。当我尝试使用nsolve查找最低根目录（这是我真正需要的唯一根目录，但是我不介意找到所有根目录时）时，出现一个错误，提示“无法从-0.499923944877944创建MPF”。

我尝试使用多个不同的求解器。当我使用SymPy的求解时，它仅找到5个根（当应该有6个时）。使用solve也花费了很长时间，因为我认为它最初是尝试以符号方式解决它。我尝试了Solveset，但没有给出正确的答案。

下面是我的所有代码。一切都会按预期进行，直到nsolve最底层为止。

from symengine import *
import sympy
from sympy import Matrix
from sympy import nsolve

trial = Matrix()

r, E1, E = symbols('r, E1, E')
H11, H22, H12, H21 = symbols("H11, H22, H12, H21")
S11, S22, S12, S21 = symbols("S11, S22, S12, S21")
low = 0
high = oo

integrate = lambda *args: sympy.N(sympy.integrate(*args))

quadratic_expression = (H11-E1*S11)*(H22-E1*S22)-(H12-E1*S12)*(H21-E1*S21)
general_solution = sympify(sympy.solve(quadratic_expression, E1)[0])


def solve_quadratic(**kwargs):
    return general_solution.subs(kwargs)


def H(fun):
    return -fun.diff(r, 2)/2 - fun.diff(r)/r - fun/r


psi0 = exp(-3*r/2)
trial = trial.row_insert(0, Matrix([psi0]))
I1 = integrate(4*pi*(r**2)*psi0*H(psi0), (r, low, high))
I2 = integrate(4*pi*(r**2)*psi0**2, (r, low, high))
E0 = I1/I2
print(E0)

for x in range(5):
    f1 = psi0
    f2 = r * (H(psi0)-E0*psi0)
    Hf1 = H(f1).simplify()
    Hf2 = H(f2).simplify()

    H11 = integrate(4*pi*(r**2)*f1*Hf1, (r, low, high))
    H12 = integrate(4*pi*(r**2)*f1*Hf2, (r, low, high))
    H21 = integrate(4*pi*(r**2)*f2*Hf1, (r, low, high))
    H22 = integrate(4*pi*(r**2)*f2*Hf2, (r, low, high))

    S11 = integrate(4*pi*(r**2)*f1**2, (r, low, high))
    S12 = integrate(4*pi*(r**2)*f1*f2, (r, low, high))
    S21 = S12
    S22 = integrate(4*pi*(r**2)*f2**2, (r, low, high))

    E0 = solve_quadratic(
            H11=H11, H22=H22, H12=H12, H21=H21,
            S11=S11, S22=S22, S12=S12, S21=S21,
        )
    print(E0)

    C = -(H11 - E0*S11)/(H12 - E0*S12)
    psi0 = (f1 + C*f2).simplify()
    trial = trial.row_insert(x+1, Matrix([[psi0]]))

# Free ICI Part

h = zeros(x+2, x+2)
HS = zeros(x+2, 1)
S = zeros(x+2, x+2)

for s in range(x+2):
    HS[s] = H(trial[s]).simplify()

for i in range(x+2):
    for j in range(x+2):
        h[i, j] = integrate(4*pi*(r**2)*trial[i]*HS[j], (r, low, high))

for i in range(x+2):
    for j in range(x+2):
        S[i, j] = integrate(4*pi*(r**2)*trial[i]*trial[j], (r, low, high))

m = h - E*S
eqn = m.det()

roots = nsolve(eqn, E0)

print(roots)

最小根应该大于或等于-0.5，但是甚至还没有达到给我根的程度。

Answer 1

使用nsolve时的最初猜测应该是浮点数，但是我输入了一个符号数。下面是更正的代码。

from symengine import *
import sympy
from sympy import Matrix
from sympy import nsolve

trial = Matrix()

r, E1, E = symbols('r, E1, E')
H11, H22, H12, H21 = symbols("H11, H22, H12, H21")
S11, S22, S12, S21 = symbols("S11, S22, S12, S21")
low = 0
high = oo

integrate = lambda *args: sympy.N(sympy.integrate(*args))

quadratic_expression = (H11-E1*S11)*(H22-E1*S22)-(H12-E1*S12)*(H21-E1*S21)
general_solution = sympify(sympy.solve(quadratic_expression, E1)[0])


def solve_quadratic(**kwargs):
    return general_solution.subs(kwargs)


def H(fun):
    return -fun.diff(r, 2)/2 - fun.diff(r)/r - fun/r


psi0 = exp(-3*r/2)
trial = trial.row_insert(0, Matrix([psi0]))
I1 = integrate(4*pi*(r**2)*psi0*H(psi0), (r, low, high))
I2 = integrate(4*pi*(r**2)*psi0**2, (r, low, high))
E0 = I1/I2
print(E0)

for x in range(6):
    f1 = psi0
    f2 = r * (H(psi0)-E0*psi0)
    Hf1 = H(f1).simplify()
    Hf2 = H(f2).simplify()

    H11 = integrate(4*pi*(r**2)*f1*Hf1, (r, low, high))
    H12 = integrate(4*pi*(r**2)*f1*Hf2, (r, low, high))
    H21 = integrate(4*pi*(r**2)*f2*Hf1, (r, low, high))
    H22 = integrate(4*pi*(r**2)*f2*Hf2, (r, low, high))

    S11 = integrate(4*pi*(r**2)*f1**2, (r, low, high))
    S12 = integrate(4*pi*(r**2)*f1*f2, (r, low, high))
    S21 = S12
    S22 = integrate(4*pi*(r**2)*f2**2, (r, low, high))

    E0 = solve_quadratic(
            H11=H11, H22=H22, H12=H12, H21=H21,
            S11=S11, S22=S22, S12=S12, S21=S21,
        )
    print(E0)

    C = -(H11 - E0*S11)/(H12 - E0*S12)
    psi0 = (f1 + C*f2).simplify()
    trial = trial.row_insert(x+1, Matrix([[psi0]]))

# Free ICI Part

h = zeros(x+2, x+2)
HS = zeros(x+2, 1)
S = zeros(x+2, x+2)

for s in range(x+2):
    HS[s] = H(trial[s]).simplify()

for i in range(x+2):
    for j in range(x+2):
        h[i, j] = integrate(4*pi*(r**2)*trial[i]*HS[j], (r, low, high))

for i in range(x+2):
    for j in range(x+2):
        S[i, j] = integrate(4*pi*(r**2)*trial[i]*trial[j], (r, low, high))

m = h - E*S
eqn = m.det()

roots = nsolve(eqn, float(E0))

print(roots)

如果有人知道如何加快nsolve的速度，我很想听听。我可能只是要为二分法编写自己的脚本，因为nsolve花费的时间太长（当我在for循环中执行11次迭代时，我一直在等待几个小时才能解决它，这很荒谬）

Answer 2

要考虑的两件事：如果您知道根的间隔，nsolve将接受一个Solver ='bisect'关键字：

>>> nsolve(x**2-3,x,(0.,2.),solver='bisect')
1.73205080756888

此外，在这种情况下，real_roots可能是获取根的有用方法。在您发布的代码中，x在range（6）范围内，则获得阶数为88的多项式，并使用以下形式对其进行了短阶求解：

>>> eq=nsimplify(eqn, rational=True).as_numer_denom()[0]
>>> [i.n(3) for i in real_roots(Poly(eq,E))][:7]
[-0.836, -0.298, -0.117, -0.0821, 0.0854, 0.181, 0.399]

为什么nsolve说“无法从-0.499923944877944创建MPF”？

2 个答案: