实数在 python 中变得复杂

Question

密码是：

A = b/a
B = c/a
C = d/a

Q = (A**2-3*B)/9
R = (2*A**3-9*A*B+27*C)/54

M = R**2-Q**3

p = (3*B-A**2)/3
q = (2*A**3-9*A*B+27*C)/27

delta = (q/2)**2+(p/3)**3

if M <= 0 :
    from math import sqrt,acos,cos,pi

    Z = acos(R/sqrt(Q**3))
    X1 = -(2*sqrt(Q)*cos(Z/3))-A/3
    X2 = -(2*sqrt(Q)*cos((Z+2*pi)/3))-A/3
    X3 = -(2*sqrt(Q)*cos((Z-2*pi)/3))-A/3


elif delta > 0 :
    from math import sqrt
    import cmath

    u = ((-q/2)+sqrt(delta))**(1/3)
    v = ((q/2)+sqrt(delta))**(1/3)

    X1 = u-v-A/3
    X2 = -(1/2)*(u-v)-A/3+(u+v)*(sqrt(3)/2)*cmath.sqrt(-1)
    X3 = -(1/2)*(u-v)-A/3-(u+v)*(sqrt(3)/2)*cmath.sqrt(-1)

当我输入时它给了我错误的答案：

a = 1 b = 1 c = -1 d = -1
M == 0

但它跳过 M 并转到 delta

结果是：

X1 = 1.0
X2 = (-1+4.8e-09j)
X3 = (-1-4.8e-09j)

我不知道怎么了？假设去 M 和 X1 ， X2 和 X3 通过在 M

中计算变为真实

Answer 1

负数的分数次幂要么给出复数（在 Python 3.x 中）（即使是奇数次幂，如 1/3 或 1/5），要么给出一个值错误Python 2.x (ValueError: negative number cannot be raised to a fractional power).

例子-

>>> (-27)**(1/3)
(1.5000000000000004+2.598076211353316j)

你需要自己处理这种情况，例如，如果你正在做 (1/3) 的幂，你将不得不像 -

x = -27
if x < 0:
    pow = -((-x) **(1/3))
elif x > 0:
    pow = x **(1/3)
else:
    pow = 0

实数在 python 中变得复杂

real numbers become complex in python

python

complex-numbers

python-3.x