I have a function as the following
q = 1 / sqrt( ((1+z)**2 * (1+0.01*o_m*z) - z*(2+z)*(1-o_m)) )
h = 5 * log10( (1+z)*q ) + 43.1601
I have experimental answers of above equation and once I must to put some data into above function and solve equation below
chi=(q_exp-q_theo)**2/err**2 # this function is a sigma, sigma chi from z=0 to z=1.4 (in the data file)
z, err and q_exp are in the data file(2.txt). Now I have to choose a range for o_m (0.2 to 0.4) and find in what o_m, the chi function will be minimized.
my code is:
from math import *
from scipy.integrate import quad
min = None
l = None
a = None
b = None
c = 0
def ant(z,om,od):
return 1/sqrt( (1+z)**2 * (1+0.01*o_m*z) - z*(2+z)*o_d )
for o_m in range(20,40,1):
o_d=1-0.01*o_m
with open('2.txt') as fp:
for line in fp:
n = list( map(float, line.split()) )
q = quad(ant,n[0],n[1],args=(o_m,o_d))[0]
h = 5.0 * log10( (1+n[1])*q ) + 43.1601
chi = (n[2]-h)**2 / n[3]**2
c = c + chi
if min is None or min>c:
min = c
l = o_m
print('chi=',q,'o_m=',0.01*l)
n[1],n[2],n[3],n[4] are z1, z2, q_exp and err, respectively in the data file. and z1 and z2 are the integration range.
I need your help and I appreciate your time and your attention.
Please do not rate a negative value. I need your answers.
Unconcsiosely, this question overlap my other question. The correct answer is:
from math import *
import numpy as np
from scipy.integrate import quad
min=l=a=b=chi=None
c=0
z,mo,err=np.genfromtxt('Union2.1_z_dm_err.txt',unpack=True)
def ant(z,o_m): #0.01*o_m is steps of o_m
return 1/sqrt(((1+z)**2*(1+0.01*o_m*z)-z*(2+z)*(1-0.01*o_m)))
for o_m in range(20,40):
c=0
for i in range(len(z)):
q=quad(ant,0,z[i],args=(o_m,))[0] #Integration o to z
h=5*log10((1+z[i])*(299000/70)*q)+25 #function of dL
chi=(mo[i]-h)**2/err[i]**2 #chi^2 test function
c=c+chi
l=o_m
print('chi^2=',c,'Om=',0.01*l,'OD=',1-0.01*l)