I have been trying to solve the following equation for R1 in python using scipy, but I get the error:
result = root_scalar(equation, bracket=[0, 8])
Traceback (most recent call last):
Cell In[1638], line 1
result = root_scalar(equation, bracket=[0, 8])
File ~\anaconda3\lib\site-packages\scipy\optimize\_root_scalar.py:275 in root_scalar
r, sol = methodc(f, a, b, args=args, **kwargs)
File ~\anaconda3\lib\site-packages\scipy\optimize\_zeros_py.py:784 in brentq
r = _zeros._brentq(f, a, b, xtol, rtol, maxiter, args, full_output, disp)
ValueError: f(a) and f(b) must have different signs
The equation is:
K= 10**(a0 + (a1 * np.log10(R1/R2)) + (a2 * (np.log10(R1/R2))**2) + (a3 * (np.log10(R1/R2))**3) + (a4 * (np.log10(R1/R2))**4)) + 0.0166
The code that I have been using is the following:
R2 = 0.002294000005349517
K = 0.09539999812841415
a0 = -1.1358
a1 = -2.1146
a2 = 1.6474
a3 = -1.1428
a4 = -0.6190
def equation(log_R1_R2):
x = log_R1_R2
return 10**(a0 + a1*x + a2*x**2 + a3*x**3 + a4*x**4) - (K - 0.0166)
result = root_scalar(equation, bracket=[0, 8])
log_R1_R2 = result.root
R1 = R2 * 10**log_R1_R2
This problem seems better suited for a polynomial root solver. Using the numpy.polynomial.Polynomial class, we can create the polynomial in question, get the roots, and find R1 using those roots. As for which roots you want to use, that's for you to decide since I don't know the application.
import numpy as np
from numpy.polynomial import Polynomial
R2 = 0.002294000005349517
K = 0.09539999812841415
a0 = -1.1358
a1 = -2.1146
a2 = 1.6474
a3 = -1.1428
a4 = -0.6190
p = Polynomial([-np.log10(K - 1/60) + a0, a1, a2, a3, a4])
roots = p.roots()
print(roots)
R1 = R2*10**roots
Output (i.e. the roots):
[-3.07249423+0.j, -0.0149377 +0.j, 0.62061419-0.86009349j, 0.62061419+0.86009349j]