Python 求两条三次曲线之间的公共切线
给定两个函数，我想整理两条曲线的公共切线： 公共切线的斜率可通过以下方式获得：Python 求两条三次曲线之间的公共切线,python,numpy,scipy,sympy,equation-solving,Python,Numpy,Scipy,Sympy,Equation Solving,给定两个函数，我想整理两条曲线的公共切线： 公共切线的斜率可通过以下方式获得： slope of common tangent = (f(x1) - g(x2)) / (x1 - x2) = f'(x1) = g'(x2) 61.6634100000000 -941.2375622594436 62.3429030000000 -941.2377748739724 62.9226515000000 -941.2378903605746 63.0043440000000 -941.237898
slope of common tangent = (f(x1) - g(x2)) / (x1 - x2) = f'(x1) = g'(x2)
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
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
f'(x1) = g'(x2) # Eq. 1
(f(x1) - g(x2)) / (x1 - x2) = f'(x1) # Eq. 2
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.datslope of common tangent = (f(x1) - g(x2)) / (x1 - x2) = f'(x1) = g'(x2)
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
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
2.datslope of common tangent = (f(x1) - g(x2)) / (x1 - x2) = f'(x1) = g'(x2)
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
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
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()
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)
slope of common tangent = (f(x1) - g(x2)) / (x1 - x2) = f'(x1) = g'(x2)
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
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
x1=61.2569899x2=59.7677843import 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()
x**5+a*x**4+…=0y=x**2x*y**2+a*y**2+…=0（立方）。我们知道这一点。所以SymPy解决不了这个问题也就不足为奇了。您需要一个数值解算器（另一个原因是Symphy并不是真正设计用来解满是浮点常量的方程的，它们对于符号操作来说很麻烦）
数值根查找
这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。这可能会使fsolve
的生活变得更轻松
范围限制：最小化
fsolve
及其相关项（如root
）都不支持在变量上设置边界。但是您可以使用最小二乘法
，它将查找表达式E1、E2的平方和的最小值。它支持上限和下限，如果幸运的话，最小值（“成本”）将在机器精度范围内为0，表示您找到了根。一个抽象的例子（因为我没有你的数据）：
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，将最小二乘法的输出分配给某个变量res
，并从那里访问res.x
。多亏了所有@if。。。。帮助，通过运行此答案末尾发布的以下代码，讨论了3条路径的结果：
1）最小平方
带默认公差：
成本为零，找到的公切线非常接近，但不理想：
2）最小平方
公差更紧：
我原以为成本会更高，但由于某种原因我不明白，成本甚至更低。找到的公共切线更接近：
3）使用fsolve
但限制区域
我们在帖子中看到，当使用x1，x2=fsolve（等式，（50，60））
时，找到的公共切线不是正确的（第二张和第三张图片）。然而，如果我们使用x1，x2=fsolve（方程式，（61.5，62））
：
我们发现，发现的斜率与最小二乘法非常相似，公差更紧。因此，找到的公共切线也非常接近：
此表总结了结果：
产生以下三种途径的代码：
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()
一个最简单的例子是从立方参数开始，如果你的问题是你认为的，我们不需要拟合过程。非常感谢你的回答。我已经实现了你的方法，但不幸的是，这并没有得到正确的等式。请参阅更新的帖子。方程式E1和E2基于P是BM的导数的假设。在您的代码中，P不是BM的导数。因此，您需要BM的导数。错误消息准确地说明了问题所在：初始点不可行（意思是：不满足约束）。第一个坐标的下限是61，但初始值是60，很接近。导数在这两点上看起来差不多，切线在另一点上也非常接近切线。似乎最小的决定它足够好，并停止。您应该阅读它的输出，看看什么终止条件是导致停止的原因，并通过减少相应的参数来收紧它。默认停止参数为ftol=1e-08、xtol=1e-08、gtol=1e-08；请参阅读取最小二乘法的完整输出也很有用，因为它包含成本，成本应该非常接近于0。如果不是，则您知道即使不绘制解决方案，该解决方案也不会很好。您没有在fsolve
中限制区域，因此没有此类选项。您提供的元组（61.5，62）
，用作起点x0（请参阅）。如果起点更接近您想要的解决方案，fsolve
更有可能找到它，而不是离开某个地方，最终收敛到另一个不想要的解决方案。