一、一元四次方程求解
一元四次方程求根公式，百科： https://baike.baidu.com/item/%E4%B8%80%E5%85%83%E5%9B%9B%E6%AC%A1%E6%96%B9%E7%A8%8B%E6%B1%82%E6%A0%B9%E5%85%AC%E5%BC%8F/10721996?fr=aladdin
1.网上搜到的求解代码
python求解代码，见：https://github.com/Larissa1990/Solve-cubic-and-quartic-equations-with-one-unknown/blob/master/Equations.py 经测试， 有bug，求解不正确
说明，见：https://www.cnblogs.com/larissa-0464/p/11706131.html
2. 一元四次方程在线求解工具：https://www.osgeo.cn/app/s2085 可用来验证求解程序是否正确；
3. 一元四次方程， 沈天珩简化求根公式, 具体公式 详见百科链接；
import math
import cmath
import numpy as np
def cal_quartic_ik(args_list):
a, b, c, d, e = args_list
D = 3*pow(b,2) - 8*a*c
E = -pow(b, 3) + 4*a*b*c - 8*pow(a, 2)*d
F = 3*pow(b, 4) + 16*pow(a, 2)*pow(c, 2) - 16*a*pow(b, 2)*c + 16*pow(a, 2)*b*d - 64*pow(a, 3)*e
A = D**2 - 3*F
B = D*F - 9*pow(E, 2)
C = F**2 - 3*D*pow(E, 2)
delta = B**2 - 4*A*C # 总判别式
if (D == 0) & (E == 0) & (F == 0):
""" 四重实根"""
x = -b/(4*a)
return 1, [x]
if (A == 0) & (B == 0) & (C == 0) & (D*E*F != 0):
""" 两个实根，其中一个三重实根"""
x1 = (-b*D + 9*E)/(4*a*D)
x234 = (-b * D - 3 * E) / (4 * a * D)
return 2, [x1, x234]
if (E == 0) & (F == 0) & (D != 0):
""" 一对二重根"""
if D>0: # 根为实数
x13 = (-b + math.sqrt(D))/(4*a)
x24 = (-b - math.sqrt(D)) / (4 * a)
return 2, [x13, x24]
if D<0: # 根为虚数
# x13 = (-b + cmath.sqrt(D))/(4*a)
# x24 = (-b - cmath.sqrt(D)) / (4 * a)
return 0, 0
if (A*B*C != 0) & (delta == 0):
""" 一对二重实根 """
x3 = (-b - np.sign(A*B*E)*math.sqrt( D - B/A))/(4*a)
x4 = (-b - np.sign(A*B*E)*math.sqrt( D - B/A))/(4*a)
if A*B>0 : # 其余两根为不等实根
x1 = (-b + np.sign(A*B*E)*math.sqrt( D - B/A) + math.sqrt( 2*B/A) )/(4*a)
x2 = (-b + np.sign(A * B * E) * math.sqrt(D - B / A) - math.sqrt(2 * B / A)) / (4 * a)
return 4, [x1, x2, x3, x4]
if A*B < 0: # 其余两根为共轭虚根
# x1 = (-b + np.sign(A * B * E) * math.sqrt(D - B / A) + cmath.sqrt(2 * B / A)) / (4 * a)
# x2 = (-b + np.sign(A * B * E) * math.sqrt(D - B / A) - cmath.sqrt(2 * B / A)) / (4 * a)
return 2, [x3, x4]
if delta > 0:
"""" 两个不等实根和一对共轭虚根"""
z1 = A*D + 3*(( -B + math.sqrt(delta))/2.0)
z2 = A * D + 3 * ((-B - math.sqrt(delta)) / 2.0)
# print """ z1 = """, z1
# print """ z2 = """, z2
# print """ abs(z1) = """, abs(z1)
# print """ abs(z2) = """, abs(z2)
z = D**2 - D*(np.sign(z1)*pow(abs(z1), 1.0/3.0) + np.sign(z2)*pow(abs(z2), 1.0/3.0)) + \
(np.sign(z1)*pow(abs(z1), 1.0/3.0) + np.sign(z2)*pow(abs(z2), 1.0/3.0))**2 - 3*A
x1 = (-b + np.sign(E)*math.sqrt((D + np.sign(z1)*pow(abs(z1), 1.0/3.0) + np.sign(z2)*pow(abs(z2), 1.0/3.0))/3.0)
+ math.sqrt((2*D - np.sign(z1)*pow(abs(z1), 1.0/3.0) - np.sign(z2)*pow(abs(z2), 1.0/3.0)
+ 2*math.sqrt(z))/3.0))/(4*a)
x2 = (-b + np.sign(E)*math.sqrt((D + np.sign(z1)*pow(abs(z1), 1.0/3.0) + np.sign(z2)*pow(abs(z2), 1.0/3.0))/3.0)
- math.sqrt((2*D - np.sign(z1)*pow(abs(z1), 1.0/3.0) - np.sign(z2)*pow(abs(z2), 1.0/3.0)
+ 2*math.sqrt(z))/3.0))/(4*a)
# 虚根忽略
return 2, [x1, x2]
if delta < 0:
if E == 0:
if (D>0) & (F>0) :
""" 四个不等实根 """
x1 = (-b + math.sqrt(D + 2*math.sqrt(F))) / (4 * a)
x2 = (-b - math.sqrt(D + 2 * math.sqrt(F))) / (4 * a)
x3 = (-b + math.sqrt(D - 2 * math.sqrt(F))) / (4 * a)
x4 = (-b - math.sqrt(D - 2 * math.sqrt(F))) / (4 * a)
return 4, [x1, x2, x3, x4]
else:
""" 两对不等共轭虚根 """
# 虚根忽略
print " 两对不等共轭虚根 "
return 0, 0
else:
if (D > 0) & (F > 0):
""" 四个不等实根 """
theta = math.acos((3*B-2*A*D)/(2*A*math.sqrt(A)))
y1 = (D - 2*math.sqrt(A)*math.cos(theta/3.0))/3.0
y2 = (D + math.sqrt(A)*(math.cos(theta/3.0) + math.sqrt(3)*math.sin(theta/3.0)))/3.0
y3 = (D + math.sqrt(A) * (math.cos(theta / 3.0) - math.sqrt(3) * math.sin(theta / 3.0))) / 3.0
x1 = (-b + np.sign(E) * math.sqrt(y1) + ( math.sqrt(y2) + math.sqrt(y3)))/(4*a)
x2 = (-b + np.sign(E) * math.sqrt(y1) - (math.sqrt(y2) + math.sqrt(y3))) / (4 * a)
x3 = (-b - np.sign(E) * math.sqrt(y1) + (math.sqrt(y2) - math.sqrt(y3))) / (4 * a)
x4 = (-b - np.sign(E) * math.sqrt(y1) - (math.sqrt(y2) - math.sqrt(y3))) / (4 * a)
return 4, [x1, x2, x3, x4]
else:
""" 两对不等共轭虚根 """
# 虚根忽略
print " 两对不等共轭虚根 "
return 0, 0
经测试，求解正确；
程序中，忽略了复数解，返回的解，均是实数解； 可作为一基础求解工具；
一、一元三次方程求解
盛金公式： https://baike.baidu.com/item/%E7%9B%9B%E9%87%91%E5%85%AC%E5%BC%8F#9
def cal_cubic_ik(args_list):
a, b, c, d = args_list
A = b**2 - 3*a*c
B = b*c - 9*a*d
C = c**2 - 3*b*d
delta = B**2 - 4*A*C # 总判别式
if (A == 0) & (B == 0):
""" 一个三重实根"""
x = -b/(3*a)
return 1, [x]
if delta > 0:
"""" 一个实根和一对共轭复根"""
y1 = A*b + 3*a*(( -B + math.sqrt(delta))/2.0)
y2 = A*b + 3*a*(( -B - math.sqrt(delta))/2.0)
print """ y1 = """, y1
print """ y2 = """, y2
print """ abs(y2) = """, abs(y2)
# 虚根忽略
x1 = (-b - (np.sign(y1)*pow(abs(y1), 1.0 / 3.0) + np.sign(y2)*pow(abs(y2), 1.0 / 3.0)) )/ (3 * a) # 负数直接开立方根会报错
return 1, [x1]
if delta == 0:
"""三个实根，其中有一个二重根 """
K = B/float(A)
x1 = -b/float(a) + K
x23 = -K/2.0
return 2, [x1, x23]
if delta < 0:
""" 三个不等实根 """
T = (2*A*b - 3*a*B)/(2*math.sqrt(pow(A, 3)))
theta = math.acos(T)
x1 = (-b - 2 * math.sqrt(A) * math.cos(theta / 3.0)) / (3 * a)
x2 = (-b + math.sqrt(A)*(math.cos(theta/3.0)) + math.sqrt(3)*math.sin(theta/3.0))/(3*a)
x3 = (-b + math.sqrt(A) * (math.cos(theta / 3.0)) - math.sqrt(3) * math.sin(theta / 3.0)) / (3 * a)
return 3, [x1, x2, x3]
程序中，忽略了复数解，返回的解，均是实数解；
来源：CSDN
作者：_yuan_
链接：https://blog.csdn.net/reasonyuanrobot/article/details/103805095