解決對於x高度非線性方程在Python

Question

我試圖解決分貝以下等式（爲簡單起見，我說分貝如問題標題X）：

所有的等式中的其他項是已知的。我嘗試使用SymPy來象徵性地解決dB問題，但我一直在獲取超時錯誤。我也嘗試使用scipy.optimize中的fminbound，但dB的答案是錯誤的（請參閱下面的Python代碼，使用fminbound方法）。

有誰知道用Python解決dB公式的方法嗎？

import numpy as np 

from scipy.optimize import fminbound 

#------------------------------------------------------------------------------ 
# parameters 

umf = 0.063   # minimum fluidization velocity, m/s 
dbed = 0.055  # bed diameter, m 
z0 = 0    # position bubbles are generated, m 
z = 0.117   # bed vertical position, m 
g = 9.81   # gravity, m/s^2 

#------------------------------------------------------------------------------ 
# calculations 

m = 3      # multiplier for Umf 
u = m*umf     # gas superficial velocity, m/s 

abed = (np.pi*dbed**2)/4.0 # bed cross-sectional area, m^2 

# calculate parameters used in equation 

dbmax = 2.59*(g**-0.2)*(abed*(u-umf))**0.4 
dbmin = 3.77*(u-umf)**2/g 

c1 = 2.56*10**-2*((dbed/g)**0.5/umf) 

c2 = (c1**2 + (4*dbmax)/dbed)**0.5 

c3 = 0.25*dbed*(c1 + c2)**2 

dbeq = 0.25*dbed*(-c1 + (c1**2 + 4*(dbmax/dbed))**0.5)**2 

# general form of equation ... (term1)^power1 * (term2)^power2 = term3 

power1 = 1 - c1/c2 

power2 = 1 + c1/c2 

term3 = np.exp(-0.3*(z - z0)/dbed) 

def dB(d): 
    term1 = (np.sqrt(d) - np.sqrt(dbeq))/(np.sqrt(dbmin) - np.sqrt(dbeq)) 
    term2 = (np.sqrt(d) + np.sqrt(c3))/(np.sqrt(dbmin) + np.sqrt(c3)) 
    return term1**power1 * term2**power2 - term3 

# solve main equation for dB 

dbub = fminbound(dB, 0.01, dbed) 

print 'dbub = ', dbub 

Answer 1

下面是四個單暗淡根的方法：

from scipy.optimize import brentq, brenth, ridder, bisect 
for rootMth in [brentq, brenth, ridder, bisect]: 
    dbub = rootMth(dB, 0.01, dbed) 
    print 'dbub = ', dbub, '; sanity check (is it a root?):', dB(dbub) 

另外牛頓 - 拉夫遜（割線/海利）方法：

from scipy.optimize import newton 
dbub = newton(dB, dbed) 
print 'dbub = ', dbub, '; sanity check (is it a root?):', dB(dbub) 

的SciPy的文檔建議brentq如果有一個包圍間隔。

Answer 2

要解決什麼在你的標題很簡單：

In [9]: 

import numpy as np 

import scipy.optimize as so 

In [10]: 

def f(x): 

    return ((x-0.32)**0.8+(x+1.45)**1.1-np.exp(0.8))**2 

In [11]: 

so.fmin(f, x0=5) 

Optimization terminated successfully. 
     Current function value: 0.000000 
     Iterations: 20 
     Function evaluations: 40 

Out[11]: 

array([ 0.45172119]) 

In [12]: 

f(0.45172119) 

Out[12]: 

4.7663411535618792e-13 

所有其它參數不變？