1） `least_squares`，默认容差为：

Question

给定两个函数，我想整理两条曲线的公共切线：

公共切线的斜率可以通过以下方式获得：

slope of common tangent = (f(x1) - g(x2)) / (x1 - x2) = f'(x1) = g'(x2)

因此，最终我们有一个包含2个未知数的2个方程组：

f'(x1) = g'(x2) # Eq. 1
(f(x1) - g(x2)) / (x1 - x2) = f'(x1) # Eq. 2

由于某些原因我不明白，python找不到解决方案：

import numpy as np
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt
import sys
from sympy import *
import sympy as sym


# Intial candidates for fit 
E0_init = -941.510817926696
V0_init = 63.54960592453
B0_init = 76.3746233515232
B0_prime_init = 4.05340727164527

# Data 1 (Red triangles): 
V_C_I, E_C_I = np.loadtxt('./1.dat', skiprows = 1).T

# Data 14 (Empty grey triangles):
V_14, E_14 = np.loadtxt('./2.dat', skiprows = 1).T

def BM(x, a, b, c, d):
        return  (2.293710449E+17)*(1E-21)* (a + b*x + c*x**2 + d*x**3 )

def P(x, b, c, d):
    return -b - 2*c*x - 3 *d*x**2

init_vals = [E0_init, V0_init, B0_init, B0_prime_init]
popt_C_I, pcov_C_I = curve_fit(BM, V_C_I, E_C_I, p0=init_vals)
popt_14, pcov_14 = curve_fit(BM, V_14, E_14, p0=init_vals)

x1 = var('x1')
x2 = var('x2')

E1 = P(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3]) - P(x2, popt_14[1], popt_14[2], popt_14[3])
print 'E1 = ', E1

E2 = ((BM(x1, popt_C_I[0], popt_C_I[1], popt_C_I[2], popt_C_I[3]) - BM(x2, popt_14[0], popt_14[1], popt_14[2], popt_14[3])) / (x1 - x2)) - P(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3])

sols = solve([E1, E2], [x1, x2])

print 'sols = ', sols

# Linspace for plotting the fitting curves:
V_C_I_lin = np.linspace(V_C_I[0], V_C_I[-1], 10000)
V_14_lin = np.linspace(V_14[0], V_14[-1], 10000)

plt.figure()
# Plotting the fitting curves:
p2, = plt.plot(V_C_I_lin, BM(V_C_I_lin, *popt_C_I), color='black', label='Cubic fit data 1' )
p6, = plt.plot(V_14_lin, BM(V_14_lin, *popt_14), 'b', label='Cubic fit data 2')

# Plotting the scattered points: 
p1 = plt.scatter(V_C_I, E_C_I, color='red', marker="^", label='Data 1', s=100)
p5 = plt.scatter(V_14, E_14, color='grey', marker="^", facecolors='none', label='Data 2', s=100)

plt.ticklabel_format(useOffset=False)
plt.show()

1.dat如下：

61.6634100000000 -941.2375622594436
62.3429030000000 -941.2377748739724
62.9226515000000 -941.2378903605746
63.0043440000000 -941.2378981684135
63.7160150000000 -941.2378864590100
64.4085050000000 -941.2377753645115
65.1046835000000 -941.2375332100225
65.8049585000000 -941.2372030376584
66.5093925000000 -941.2367456992965
67.2180970000000 -941.2361992239395
67.9311515000000 -941.2355493856510

2.dat如下：

54.6569312500000 -941.2300821583739
55.3555152500000 -941.2312112888004
56.1392347500000 -941.2326135552780
56.9291575000000 -941.2338291772218
57.6992532500000 -941.2348135408652
58.4711572500000 -941.2356230099117
59.2666985000000 -941.2362715934311
60.0547935000000 -941.2367074271724
60.8626545000000 -941.2370273047416

更新：使用@if ....方法：

import numpy as np
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt
from matplotlib.font_manager import FontProperties

# Intial candidates for fit, per FU: - thus, the E vs V input data has to be per FU
E0_init = -941.510817926696   
V0_init = 63.54960592453  
B0_init = 76.3746233515232  
B0_prime_init = 4.05340727164527 

def BM(x, a, b, c, d):
         return  a + b*x + c*x**2 + d*x**3 

def devBM(x, b, c, d):
         return  b + 2*c*x + 3*d*x**2 

# Data 1 (Red triangles): 
V_C_I, E_C_I = np.loadtxt('./1.dat', skiprows = 1).T

# Data 14 (Empty grey triangles):
V_14, E_14 = np.loadtxt('./2.dat', skiprows = 1).T

init_vals = [E0_init, V0_init, B0_init, B0_prime_init]
popt_C_I, pcov_C_I = curve_fit(BM, V_C_I, E_C_I, p0=init_vals)
popt_14, pcov_14 = curve_fit(BM, V_14, E_14, p0=init_vals)

from scipy.optimize import fsolve
def equations(p):
    x1, x2 = p
    E1 = devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3]) - devBM(x2, popt_14[1], popt_14[2], popt_14[3])
    E2 = ((BM(x1, popt_C_I[0], popt_C_I[1], popt_C_I[2], popt_C_I[3]) - BM(x2, popt_14[0], popt_14[1], popt_14[2], popt_14[3])) / (x1 - x2)) - devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3])
    return (E1, E2)

x1, x2 =  fsolve(equations, (50, 60))
print 'x1 = ', x1
print 'x2 = ', x2

slope_common_tangent = devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3])
print 'slope_common_tangent = ', slope_common_tangent

def comm_tangent(x, x1, slope_common_tangent):
   return BM(x1, popt_C_I[0], popt_C_I[1], popt_C_I[2], popt_C_I[3]) - slope_common_tangent * x1 + slope_common_tangent * x

# Linspace for plotting the fitting curves:
V_C_I_lin = np.linspace(V_C_I[0], V_C_I[-1], 10000)
V_14_lin = np.linspace(V_14[0], V_14[-1], 10000)

plt.figure()

# Plotting the fitting curves:
p2, = plt.plot(V_C_I_lin, BM(V_C_I_lin, *popt_C_I), color='black', label='Cubic fit Calcite I' )
p6, = plt.plot(V_14_lin, BM(V_14_lin, *popt_14), 'b', label='Cubic fit Calcite II')

xp = np.linspace(54, 68, 100)
pcomm_tangent, = plt.plot(xp, comm_tangent(xp, x1, slope_common_tangent), 'green', label='Common tangent')

# Plotting the scattered points: 
p1 = plt.scatter(V_C_I, E_C_I, color='red', marker="^", label='Calcite I', s=100)
p5 = plt.scatter(V_14, E_14, color='grey', marker="^", facecolors='none', label='Calcite II', s=100)

fontP = FontProperties()
fontP.set_size('13')

plt.legend((p1, p2, p5, p6, pcomm_tangent), ("1", "Cubic fit 1", "2", 'Cubic fit 2', 'Common tangent'), prop=fontP)

print 'devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3]) = ', devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3])

plt.ylim(-941.240, -941.225) 
plt.ticklabel_format(useOffset=False)

plt.show()

我能够找到公共切线，如下所示：

但是，此公共切线对应于数据范围之外的区域中的公共切线，即使用

V_C_I_lin = np.linspace(V_C_I[0]-30, V_C_I[-1], 10000)
V_14_lin = np.linspace(V_14[0]-20, V_14[-1]+2, 10000)
xp = np.linspace(40, 70, 100)
plt.ylim(-941.25, -941.18)

可以看到以下内容：

是否可以将求解器约束到我们有数据的范围，以便找到所需的公共切线？

更新2.1 ：使用@if ....范围限制方法，以下代码会生成x1 = 61.2569899和x2 = 59.7677843。如果我们绘制它：

import numpy as np
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt
from matplotlib.font_manager import FontProperties
import sys
from sympy import *
import sympy as sym
import os

# Intial candidates for fit, per FU: - thus, the E vs V input data has to be per FU
E0_init = -941.510817926696  # -1882.50963222/2.0 
V0_init = 63.54960592453 #125.8532/2.0 
B0_init = 76.3746233515232 #74.49 
B0_prime_init = 4.05340727164527 #4.15

def BM(x, a, b, c, d):
         return  a + b*x + c*x**2 + d*x**3

def devBM(x, b, c, d):
         return  b + 2*c*x + 3*d*x**2

# Data 1 (Red triangles): 
V_C_I, E_C_I = np.loadtxt('./1.dat', skiprows = 1).T

# Data 14 (Empty grey triangles):
V_14, E_14 = np.loadtxt('./2.dat', skiprows = 1).T

init_vals = [E0_init, V0_init, B0_init, B0_prime_init]
popt_C_I, pcov_C_I = curve_fit(BM, V_C_I, E_C_I, p0=init_vals)
popt_14, pcov_14 = curve_fit(BM, V_14, E_14, p0=init_vals)

def equations(p):
    x1, x2 = p
    E1 = devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3]) - devBM(x2, popt_14[1], popt_14[2], popt_14[3])
    E2 = ((BM(x1, popt_C_I[0], popt_C_I[1], popt_C_I[2], popt_C_I[3]) - BM(x2, popt_14[0], popt_14[1], popt_14[2], popt_14[3])) / (x1 - x2)) - devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3])
    return (E1, E2)

from scipy.optimize import least_squares
lb = (61.0, 59.0)   # lower bounds on x1, x2
ub = (62.0, 60.0)    # upper bounds
result = least_squares(equations, [61, 59], bounds=(lb, ub))
print 'result = ', result

# The result obtained is:
# x1 = 61.2569899
# x2 = 59.7677843

slope_common_tangent = devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3])
print 'slope_common_tangent = ', slope_common_tangent


def comm_tangent(x, x1, slope_common_tangent):
   return BM(x1, popt_C_I[0], popt_C_I[1], popt_C_I[2], popt_C_I[3]) - slope_common_tangent * x1 + slope_common_tangent * x

# Linspace for plotting the fitting curves:
V_C_I_lin = np.linspace(V_C_I[0]-2, V_C_I[-1], 10000)
V_14_lin = np.linspace(V_14[0], V_14[-1]+2, 10000)


fig_handle = plt.figure()

# Plotting the fitting curves:
p2, = plt.plot(V_C_I_lin, BM(V_C_I_lin, *popt_C_I), color='black' )
p6, = plt.plot(V_14_lin, BM(V_14_lin, *popt_14), 'b' )

xp = np.linspace(54, 68, 100)
pcomm_tangent, = plt.plot(xp, comm_tangent(xp, x1, slope_common_tangent), 'green', label='Common tangent')

# Plotting the scattered points: 
p1 = plt.scatter(V_C_I, E_C_I, color='red', marker="^", label='1', s=100)
p5 = plt.scatter(V_14, E_14, color='grey', marker="^", facecolors='none', label='2', s=100)

fontP = FontProperties()
fontP.set_size('13')

plt.legend((p1, p2, p5, p6, pcomm_tangent), ("1", "Cubic fit 1", "2", 'Cubic fit 2', 'Common tangent'), prop=fontP)


plt.ticklabel_format(useOffset=False)

plt.show()

我们发现我们没有获得共同的切线：

Answer 1

符号根发现

您的方程组由二次方程和三次方程组成。这种系统没有封闭形式的符号解决方案。实际上，如果存在，则可以通过引入x**5 + a*x**4 + ... = 0（二次）并将原始方程重写为y = x**2（立方）来将其应用于一般的5度方程x*y**2 + a*y**2 + ... = 0。我们知道can't be done。所以SymPy无法解决它并不奇怪。你需要一个数值求解器（另一个原因是SymPy并不是真的设计用来解决充满浮点常数的方程，它们对符号操作很麻烦）。

数字根发现

SciPy fsolve是我想到的第一件事。你可以这样做：

def F(x):
    x1, x2 = x[0], x[1]
    E1 = P(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3]) - P(x2, popt_14[1], popt_14[2], popt_14[3])
    E2 = ((BM(x1, popt_C_I[0], popt_C_I[1], popt_C_I[2], popt_C_I[3]) - BM(x2, popt_14[0], popt_14[1], popt_14[2], popt_14[3])) / (x1 - x2)) - P(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3])
    return [E1, E2] 

print fsolve(F, [50, 60])    # some reasonable initial point

顺便说一下，我会从E2中的分母移动（x1-x2），重写E2为

(...) - (x1 - x2) * P(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3])

所以系统是多项式的。这可能会使fsolve的生命变得更容易。

范围限制：最小化

fsolve及其root之类的亲戚都不支持在变量上放置边界。但是你可以使用least_squares来查找表达式E1，E2的平方和的最小值。它支持上限和下限，运气好的话，最小值（“cost”）在机器精度范围内为0，表示你找到了一个根。一个抽象的例子（因为我没有你的数据）：

f1 = lambda x: 2*x**3 + 20
df1 = lambda x: 6*x**2   # derivative of f1. 
f2 = lambda x: (x-3)**3 + x
df2 = lambda x: 3*(x-3)**2 + 1

def eqns(x):
    x1, x2 = x[0], x[1]
    eq1 = df1(x1) - df2(x2)
    eq2 = df1(x1)*(x1 - x2) - (f1(x1) - f2(x2))
    return [eq1, eq2]

from scipy.optimize import least_squares
lb = (2, -2)   # lower bounds on x1, x2
ub = (5, 3)    # upper bounds
least_squares(eqns, [3, 1], bounds=(lb, ub))

输出：

 active_mask: array([0, 0])
        cost: 2.524354896707238e-29
         fun: array([7.10542736e-15, 0.00000000e+00])
        grad: array([1.93525199e-13, 1.34611132e-13])
         jac: array([[27.23625045, 18.94483256],
       [66.10672633, -0.        ]])
     message: '`gtol` termination condition is satisfied.'
        nfev: 8
        njev: 8
  optimality: 2.4802477446153134e-13
      status: 1
     success: True
           x: array([ 2.26968753, -0.15747203])

成本非常小，所以我们有一个解决方案，它是x。通常情况下，会将least_squares的输出分配给某个变量res并从那里访问res.x。

Answer 2

感谢所有@if ....帮助，通过运行本答案末尾发布的以下代码，讨论了3个途径的结果：

1） `least_squares`，默认容差为：

####  ftol=1e-08, xtol=1e-08, gtol=1e-08  #####

result =   active_mask: array([0, 0])
        cost: 4.7190963603923405e-10
         fun: array([  1.60076424e-05,   2.62216448e-05])
        grad: array([ -3.22762954e-09,  -4.72137063e-09])
         jac: array([[  2.70753295e-04,  -3.41257603e-04],
       [ -2.88378229e-04,   2.82727898e-05]])
     message: '`gtol` termination condition is satisfied.'
        nfev: 4
        njev: 4
  optimality: 2.398161337354571e-09
      status: 1
     success: True
           x: array([ 61.2569899,  59.7677843])
result.x =  [ 61.2569899  59.7677843]



slope_common_tangent =  -0.000533908881355

成本为零，找到的公共切线非常接近，但并不理想：

公差较小的

2） `least_squares`：

####  ftol=1e-12, xtol=1e-12, gtol=1e-12  #####

result_tight_tols =   active_mask: array([0, 0])
        cost: 4.3861335617475759e-20
         fun: array([  2.08437035e-10,   2.10420231e-10])
        grad: array([ -5.40980234e-16,  -7.19344843e-14])
         jac: array([[  2.69431398e-04,  -3.45167744e-04],
       [ -2.69462978e-04,   5.34965061e-08]])
     message: '`gtol` termination condition is satisfied.'
        nfev: 8
        njev: 8
  optimality: 7.6628451334260651e-15
      status: 1
     success: True
           x: array([ 61.35744106,  59.89347466])
result_tight_tols.x =  [ 61.35744106  59.89347466]



slope_common_tangent =  -0.000506777791299

我原本期望成本更高，但由于某些原因我不明白，成本更低。找到的常见切线更接近：

3）使用`fsolve`但限制区域

我们在帖子中看到，当使用x1, x2 = fsolve(equations, (50, 60))时，找到的公共切线不正确（第2张和第3张图像）。但是，如果我们使用x1, x2 = fsolve(equations, (61.5, 62))：

#### Using `fsolve`, but restricting the region:  ####


x1 =  61.3574418449
x2 =  59.8934758773
slope_common_tangent =  -0.000506777580856

我们发现找到的斜率与least_squares非常相似，且公差更小。因此，找到的共同切线也非常接近：

此表总结了结果：

产生这三种途径的代码：

import numpy as np
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt
from matplotlib.font_manager import FontProperties
import sys
from sympy import *
import sympy as sym
import os
import pickle as pl


# Intial candidates for fit, per FU: - thus, the E vs V input data has to be per FU
E0_init = -941.510817926696  # -1882.50963222/2.0 
V0_init = 63.54960592453 #125.8532/2.0 
B0_init = 76.3746233515232 #74.49 
B0_prime_init = 4.05340727164527 #4.15

def BM(x, a, b, c, d):
         return  a + b*x + c*x**2 + d*x**3 

def devBM(x, b, c, d):
         return  b + 2*c*x + 3*d*x**2 

# Data 1 (Red triangles): 
V_C_I, E_C_I = np.loadtxt('./1.dat', skiprows = 1).T

# Data 14 (Empty grey triangles):
V_14, E_14 = np.loadtxt('./2.dat', skiprows = 1).T

init_vals = [E0_init, V0_init, B0_init, B0_prime_init]
popt_C_I, pcov_C_I = curve_fit(BM, V_C_I, E_C_I, p0=init_vals)
popt_14, pcov_14 = curve_fit(BM, V_14, E_14, p0=init_vals)


def equations(p):
    x1, x2 = p
    E1 = devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3]) - devBM(x2, popt_14[1], popt_14[2], popt_14[3])
    E2 = ((BM(x1, popt_C_I[0], popt_C_I[1], popt_C_I[2], popt_C_I[3]) - BM(x2, popt_14[0], popt_14[1], popt_14[2], popt_14[3])) / (x1 - x2)) - devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3])
    return (E1, E2)

from scipy.optimize import least_squares
lb = (61.0, 59.0)   # lower bounds on x1, x2
ub = (62.0, 60.0)   # upper bounds
result = least_squares(equations, [61, 59], bounds=(lb, ub))
result_tight_tols = least_squares(equations, [61, 59], ftol=1e-12, xtol=1e-12, gtol=1e-12, bounds=(lb, ub))

print """
####  ftol=1e-08, xtol=1e-08, gtol=1e-08  #####
"""
print 'result = ', result
print 'result.x = ', result.x
print """

"""
x1 = result.x[0]
x2 = result.x[1]

slope_common_tangent = devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3])
print 'slope_common_tangent = ', slope_common_tangent

def comm_tangent(x, x1, slope_common_tangent):
   return BM(x1, popt_C_I[0], popt_C_I[1], popt_C_I[2], popt_C_I[3]) - slope_common_tangent * x1 + slope_common_tangent * x

# Linspace for plotting the fitting curves:
V_C_I_lin = np.linspace(V_C_I[0]-2, V_C_I[-1], 10000)
V_14_lin = np.linspace(V_14[0], V_14[-1]+2, 10000)


plt.figure()

# Plotting the fitting curves:
p2, = plt.plot(V_C_I_lin, BM(V_C_I_lin, *popt_C_I), color='black' )
p6, = plt.plot(V_14_lin, BM(V_14_lin, *popt_14), 'b' )

xp = np.linspace(54, 68, 100)
pcomm_tangent, = plt.plot(xp, comm_tangent(xp, x1, slope_common_tangent), 'green', label='Common tangent')

# Plotting the scattered points: 
p1 = plt.scatter(V_C_I, E_C_I, color='red', marker="^", label='1', s=100)
p5 = plt.scatter(V_14, E_14, color='grey', marker="^", facecolors='none', label='2', s=100)

fontP = FontProperties()
fontP.set_size('13')

plt.legend((p1, p2, p5, p6, pcomm_tangent), ("1", "Cubic fit 1", "2", 'Cubic fit 2', 'Common tangent'), prop=fontP)
plt.title('Least squares. Default tolerances: ftol=1e-08, xtol=1e-08, gtol=1e-08')

plt.ticklabel_format(useOffset=False)

### Tighter tolerances:
print """
####  ftol=1e-12, xtol=1e-12, gtol=1e-12  #####
"""
print 'result_tight_tols = ', result_tight_tols
print 'result_tight_tols.x = ', result_tight_tols.x
print """

"""
x1 = result_tight_tols.x[0]
x2 = result_tight_tols.x[1]

slope_common_tangent = devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3])
print 'slope_common_tangent = ', slope_common_tangent

def comm_tangent(x, x1, slope_common_tangent):
   return BM(x1, popt_C_I[0], popt_C_I[1], popt_C_I[2], popt_C_I[3]) - slope_common_tangent * x1 + slope_common_tangent * x

# Linspace for plotting the fitting curves:
V_C_I_lin = np.linspace(V_C_I[0]-2, V_C_I[-1], 10000)
V_14_lin = np.linspace(V_14[0], V_14[-1]+2, 10000)


plt.figure()

# Plotting the fitting curves:
p2, = plt.plot(V_C_I_lin, BM(V_C_I_lin, *popt_C_I), color='black' )
p6, = plt.plot(V_14_lin, BM(V_14_lin, *popt_14), 'b' )

xp = np.linspace(54, 68, 100)
pcomm_tangent, = plt.plot(xp, comm_tangent(xp, x1, slope_common_tangent), 'green', label='Common tangent')

# Plotting the scattered points: 
p1 = plt.scatter(V_C_I, E_C_I, color='red', marker="^", label='1', s=100)
p5 = plt.scatter(V_14, E_14, color='grey', marker="^", facecolors='none', label='2', s=100)

fontP = FontProperties()
fontP.set_size('13')

plt.legend((p1, p2, p5, p6, pcomm_tangent), ("1", "Cubic fit 1", "2", 'Cubic fit 2', 'Common tangent'), prop=fontP)
plt.title('ftol=1e-08, xtol=1e-08, gtol=1e-08')

plt.ticklabel_format(useOffset=False)

plt.title('Lest Squares. Tightening tolerances: ftol=1e-12, xtol=1e-12, gtol=1e-12')

print """
#### Using `fsolve`, but restricting the region:  ####

"""

from scipy.optimize import fsolve
x1, x2 =  fsolve(equations, (61.5, 62))

print 'x1 = ', x1
print 'x2 = ', x2

slope_common_tangent = devBM(x1, popt_C_I[1], popt_C_I[2], popt_C_I[3])
print 'slope_common_tangent = ', slope_common_tangent

plt.figure()

# Plotting the fitting curves:
p2, = plt.plot(V_C_I_lin, BM(V_C_I_lin, *popt_C_I), color='black' )
p6, = plt.plot(V_14_lin, BM(V_14_lin, *popt_14), 'b' )

xp = np.linspace(54, 68, 100)
pcomm_tangent, = plt.plot(xp, comm_tangent(xp, x1, slope_common_tangent), 'green', label='Common tangent')

# Plotting the scattered points: 
p1 = plt.scatter(V_C_I, E_C_I, color='red', marker="^", label='1', s=100)
p5 = plt.scatter(V_14, E_14, color='grey', marker="^", facecolors='none', label='2', s=100)

fontP = FontProperties()
fontP.set_size('13')

plt.legend((p1, p2, p5, p6, pcomm_tangent), ("1", "Cubic fit 1", "2", 'Cubic fit 2', 'Common tangent'), prop=fontP)
plt.ticklabel_format(useOffset=False)

plt.title('Using `fsolve`, but restricting the region')



plt.show()

找到两条三次曲线之间的公共切线

2 个答案:

符号根发现

数字根发现

范围限制：最小化

1） `least_squares`，默认容差为：

2） `least_squares`：

3）使用`fsolve`但限制区域

找到两条三次曲线之间的公共切线

2 个答案:

符号根发现

数字根发现

范围限制：最小化

1） least_squares，默认容差为：

2） least_squares：

3）使用fsolve但限制区域

1） `least_squares`，默认容差为：

2） `least_squares`：

3）使用`fsolve`但限制区域