找到两条三次曲线之间的公共切线
Find common tangent line between two cubic curves
给定两个函数，我想找出两条曲线的公切线：
公切线的斜率可以通过以下方式得到：
slope of common tangent = (f(x1) - g(x2)) / (x1 - x2) = f'(x1) = g'(x2)
所以最后我们有一个包含 2 个方程和 2 个未知数的系统：
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没有找到解决方案：
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()
我能找到公切线，如下图：
但是这个公切线对应的是数据范围外区域的公切线，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：
是否可以将求解器限制在我们拥有数据的范围内，以便找到所需的公切线？
更新 2.1：使用 @if.... 范围约束方法，以下代码产生 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()
我们看到我们没有获得公切线：
求符号根
你的方程组由一个二次方程和一个三次方程组成。这样的系统没有封闭形式的符号解。事实上，如果有的话，人们可以通过引入 y = x**2（二次）并将原始方程重写为 x*y**2 + a*y**2 + ... = 0（三次），将其应用于一般的 5 次方程 x**5 + a*x**4 + ... = 0。我们知道 can't be done。所以 SymPy 无法解决也就不足为奇了。您需要一个数值求解器（另一个原因是 SymPy 并不是真正设计用于求解充满浮点常数的方程式，它们对于符号操作来说很麻烦）。
求数根
SciPy fsolve 是第一个想到的。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，我会将 (x1-x2) 从 E2 中的分母移开，将 E2 重写为
(...) - (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。这可能会让 fsolve 的生活更轻松一些。
范围限制：最小化
fsolve 及其类似 root 都不支持对变量设置界限。但是您可以使用 least_squares 来寻找表达式 E1、E2 的平方和的最小值。它支持上限和下限，如果运气好的话，最小值（"cost"）将在机器精度内为0，表示您找到了一个根。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])
成本很小，所以我们有办法，就是x。通常，将 least_squares 的输出分配给某个变量 res 并从那里访问 res.x。
感谢所有@if....的帮助，通过运行下面的代码post编辑在这个答案的末尾，讨论了3个路径的结果：
1) least_squares 默认公差：
#### 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，找到的公切线很接近，但不理想：
2) least_squares 公差更小：
#### 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
我原以为成本会更高，但不知为何，成本更低。找到的公切线更接近：
3)使用fsolve但限制区域
我们在 post 中看到，当使用 x1, x2 = fsolve(equations, (50, 60)) 时，找到的公切线不是正确的（第二和第三张图像）。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
我们看到找到的斜率与 least_squares 非常相似，公差更小。这样，找到的公切线也很接近：
这个table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()
给定两个函数，我想找出两条曲线的公切线：
公切线的斜率可以通过以下方式得到：
slope of common tangent = (f(x1) - g(x2)) / (x1 - x2) = f'(x1) = g'(x2)
所以最后我们有一个包含 2 个方程和 2 个未知数的系统：
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没有找到解决方案：
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()
我能找到公切线，如下图：
但是这个公切线对应的是数据范围外区域的公切线，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：
是否可以将求解器限制在我们拥有数据的范围内，以便找到所需的公切线？
更新 2.1：使用 @if.... 范围约束方法，以下代码产生 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()
我们看到我们没有获得公切线：
求符号根
你的方程组由一个二次方程和一个三次方程组成。这样的系统没有封闭形式的符号解。事实上，如果有的话，人们可以通过引入 y = x**2（二次）并将原始方程重写为 x*y**2 + a*y**2 + ... = 0（三次），将其应用于一般的 5 次方程 x**5 + a*x**4 + ... = 0。我们知道 can't be done。所以 SymPy 无法解决也就不足为奇了。您需要一个数值求解器（另一个原因是 SymPy 并不是真正设计用于求解充满浮点常数的方程式，它们对于符号操作来说很麻烦）。
求数根
SciPy fsolve 是第一个想到的。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，我会将 (x1-x2) 从 E2 中的分母移开，将 E2 重写为
(...) - (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。这可能会让 fsolve 的生活更轻松一些。
范围限制：最小化
fsolve 及其类似 root 都不支持对变量设置界限。但是您可以使用 least_squares 来寻找表达式 E1、E2 的平方和的最小值。它支持上限和下限，如果运气好的话，最小值（"cost"）将在机器精度内为0，表示您找到了一个根。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])
成本很小，所以我们有办法，就是x。通常，将 least_squares 的输出分配给某个变量 res 并从那里访问 res.x。
感谢所有@if....的帮助，通过运行下面的代码post编辑在这个答案的末尾，讨论了3个路径的结果：
1) least_squares 默认公差：
#### 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，找到的公切线很接近，但不理想：
2) least_squares 公差更小：
#### 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
我原以为成本会更高，但不知为何，成本更低。找到的公切线更接近：
3)使用fsolve但限制区域
我们在 post 中看到，当使用 x1, x2 = fsolve(equations, (50, 60)) 时，找到的公切线不是正确的（第二和第三张图像）。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
我们看到找到的斜率与 least_squares 非常相似，公差更小。这样，找到的公切线也很接近：
这个table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()