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python solve Cubic equation without using sympy

Is it possible to solve Cubic equation without using sympy?
Example:
import sympy as sp
xp = 30
num = xp + 4.44
sp.var('x, a, b, c, d')
Sol3 = sp.solve(0.0509 * x ** 3 + 0.0192 * x ** 2 + 3.68 * x - num, x)
The result is:
[6.07118098358257, -3.2241955998463 - 10.0524891203436*I, -3.2241955998463 + 10.0524891203436*I]
But I want to find a way to do it with numpy or without 3 part lib at all
I tried with numpy:
import numpy as np
coeff = [0.0509, 0.0192, 3.68, --4.44]
print(np.roots(coeff))
But the result is :
[ 0.40668245+8.54994773j 0.40668245-8.54994773j -1.19057511+0.j]

In your numpy method you are making two slight mistakes with the final coefficient.
In the SymPy example your last coefficient is - num, this is, according to your code: -num = - (xp + 4.44) = -(30 + 4.44) = -34.44
In your NumPy example yout last coefficient is --4.44, which is 4.44 and does not equal -34.33.
If you edit the NumPy code you will get:
import numpy as np
coeff = [0.0509, 0.0192, 3.68, -34.44]
print(np.roots(coeff))
[-3.2241956 +10.05248912j -3.2241956 -10.05248912j
6.07118098 +0.j ]
The answer are thus the same (note that NumPy uses j to indicate a complex number. SymPy used I)

You could implement the cubic formula
this Youtube video from mathologer could help understand it.
Based on that, the cubic function for ax^3 + bx^2 + cx + d = 0 can be written like this:
def cubic(a,b,c,d):
n = -b**3/27/a**3 + b*c/6/a**2 - d/2/a
s = (n**2 + (c/3/a - b**2/9/a**2)**3)**0.5
r0 = (n-s)**(1/3)+(n+s)**(1/3) - b/3/a
r1 = (n+s)**(1/3)+(n+s)**(1/3) - b/3/a
r2 = (n-s)**(1/3)+(n-s)**(1/3) - b/3/a
return (r0,r1,r2)
The simplified version of the formula only needs to get c and d as parameters (aka p and q) and can be implemented like this:
def cubic(p,q):
n = -q/2
s = (q*q/4+p**3/27)**0.5
r0 = (n-s)**(1/3)+(n+s)**(1/3)
r1 = (n+s)**(1/3)+(n+s)**(1/3)
r2 = (n-s)**(1/3)+(n-s)**(1/3)
return (r0,r1,r2)
print(cubic(-15,-126))
(5.999999999999999, 9.999999999999998, 2.0)
I'll let you mix in complex number operations to properly get all 3 roots

Runtime error: Factor is exactly singular

I am trying to implement 2 temperature models, the following equations:
C_e(∂T_e)/∂t=∇[k_e∇T_e ]-G(T_e-T_ph )+ A(r,t)
C_ph(∂T_ph)/∂t=∇[k_ph∇T_ph] + G(T_e-T_ph)
Code
from fipy.tools import numerix
import scipy
import fipy
import numpy as np
from fipy import CylindricalGrid1D
from fipy import Variable, CellVariable, TransientTerm, DiffusionTerm, Viewer, LinearLUSolver, LinearPCGSolver, \
LinearGMRESSolver, ImplicitDiffusionTerm, Grid1D
FIPY_SOLVERS = scipy
## Mesh
nr = 50
dr = 1e-7
# r = nr * dr
mesh = CylindricalGrid1D(nr=nr, dr=dr, origin=0)
x = mesh.cellCenters[0]
# Variables
T_e = CellVariable(name="electronTemp", mesh=mesh,hasOld=True)
T_e.setValue(300)
T_ph = CellVariable(name="phononTemp", mesh=mesh, hasOld=True)
T_ph.setValue(300)
G = CellVariable(name="EPC", mesh=mesh)
t = Variable()
# Material parameters
C_e = CellVariable(name="C_e", mesh=mesh)
k_e = CellVariable(name="k_e", mesh=mesh)
C_ph = CellVariable(name="C_ph", mesh=mesh)
k_ph = CellVariable(name="k_ph", mesh=mesh)
C_e = 4.15303 - (4.06897 * numerix.exp(T_e / -85120.8644))
C_ph = 4.10446 - 3.886 * numerix.exp(-T_ph / 373.8)
k_e = 0.1549 * T_e**-0.052
k_ph =1.24 + 16.29 * numerix.exp(-T_ph / 151.57)
G = numerix.exp(21.87 + 10.062 * numerix.log(numerix.log(T_e )- 5.4))
# Boundary conditions
T_e.constrain(300, where=x > 4.5e-6)
T_ph.constrain(300, where=x > 4.5e-6)
# Source 𝐴(𝑟,𝑡) = 𝑎𝐷(𝑟)𝜏−1 𝑒−𝑡/𝜏 , 𝐷(𝑟) = 𝑆𝑒 exp (−𝑟2/𝜎2)/√2𝜋𝜎2
sig = 1.0e-6
tau = 1e-15
S_e = 35
d_r = (S_e * 1.6e-9 * numerix.exp(-x**2 /sig**2)) / (numerix.sqrt(2. * 3.14 * sig**2))
A_t = numerix.exp(-t/tau)
a = (numerix.sqrt(2. * 3.14)) / (3.14 * sig)
A_r = a * d_r * tau**-1 * A_t
eq0 = (TransientTerm(var=T_e, coeff=C_e) == DiffusionTerm(var=T_e, coeff=k_e) - G*(T_e - T_ph) + A_r
eq1 =(TransientTerm(var=T_ph, coeff=C_ph) == DiffusionTerm(var=T_ph, coeff=k_ph) + G*(T_e - T_ph)
eq = eq0 & eq1
dt = 1e-18
steps = 7000
elapsed = 0.
vi = Viewer((T_e, T_ph), datamin=0., datamax=2e4)
for step in range(steps):
T_e.updateOld()
T_ph.updateOld()
vi.plot()
res = 1e100
dt *= 1.1
while res > 1:
res = eq.sweep(dt=dt)
print(t, res)
t.setValue(t + dt)
Problem
The code is working fine with very small dt = 1e-18, but I need to run it until e 1e-10.
With this time step is going to take very long time, and setting dt *= 1.1 the resduals at some point start to increase then
gives following runtime error:
factor is exactly singular
Even with very small increment dt*= 1.005 the same issue pop up.
Using dt= 1.001 runs the code for quit long time then the residual get stuck at certain value.
Questions
I there any error in the fipy formalism of the equations?
What causes the error?
Is the error because of time step increase? If yes, how can I increase my time step?

I've made a few more changes to the code that can get you to an elapsed time of 1e-10. The main changes are
Using ImplicitSourceTerm for the terms with G. This stabalizes the solution.
Applied underRelaxation=0.5 in the sweep step. This slows down the updates in the sweep loop so the feedback loop is damped down.
Removed FIPY_SOLVERS=scipy. This isn't doing anything. FIPY_SOLVERS is an environment variable that you set outside of the Python environment.
The way the boundary conditions were applied seemed strange so I applied them in a more canonical way.
The sweep loop is fixed to 10 sweeps to get to a steady state quickly. Note that as the solution gets close to a stable steady state, the residual won't get better necessarily. Probably want to go back to residual checks if you need an accurate transient.
from fipy.tools import numerix
import scipy
import fipy
import numpy as np
from fipy import CylindricalGrid1D
from fipy import Variable, CellVariable, TransientTerm, DiffusionTerm, Viewer, LinearLUSolver, LinearPCGSolver, \
LinearGMRESSolver, ImplicitDiffusionTerm, Grid1D, ImplicitSourceTerm
## Mesh
nr = 50
dr = 1e-7
# r = nr * dr
mesh = CylindricalGrid1D(nr=nr, dr=dr, origin=0)
x = mesh.cellCenters[0]
# Variables
T_e = CellVariable(name="electronTemp", mesh=mesh,hasOld=True)
T_e.setValue(300)
T_ph = CellVariable(name="phononTemp", mesh=mesh, hasOld=True)
T_ph.setValue(300)
G = CellVariable(name="EPC", mesh=mesh)
t = Variable()
# Material parameters
C_e = CellVariable(name="C_e", mesh=mesh)
k_e = CellVariable(name="k_e", mesh=mesh)
C_ph = CellVariable(name="C_ph", mesh=mesh)
k_ph = CellVariable(name="k_ph", mesh=mesh)
C_e = 4.15303 - (4.06897 * numerix.exp(T_e / -85120.8644))
C_ph = 4.10446 - 3.886 * numerix.exp(-T_ph / 373.8)
k_e = 0.1549 * T_e**-0.052
k_ph =1.24 + 16.29 * numerix.exp(-T_ph / 151.57)
G = numerix.exp(21.87 + 10.062 * numerix.log(numerix.log(T_e )- 5.4))
# Boundary conditions
T_e.constrain(300, where=mesh.facesRight)
T_ph.constrain(300, where=mesh.facesRight)
# Source 𝐴(𝑟,𝑡) = 𝑎𝐷(𝑟)𝜏−1 𝑒−𝑡/𝜏 , 𝐷(𝑟) = 𝑆𝑒 exp (−𝑟2/𝜎2)/√2𝜋𝜎2
sig = 1.0e-6
tau = 1e-15
S_e = 35
d_r = (S_e * 1.6e-9 * numerix.exp(-x**2 /sig**2)) / (numerix.sqrt(2. * 3.14 * sig**2))
A_t = numerix.exp(-t/tau)
a = (numerix.sqrt(2. * 3.14)) / (3.14 * sig)
A_r = a * d_r * tau**-1 * A_t
eq0 = (
TransientTerm(var=T_e, coeff=C_e) == \
DiffusionTerm(var=T_e, coeff=k_e) - \
ImplicitSourceTerm(coeff=G, var=T_e) + \
ImplicitSourceTerm(var=T_ph, coeff=G) + \
A_r)
eq1 = (TransientTerm(var=T_ph, coeff=C_ph) == DiffusionTerm(var=T_ph, coeff=k_ph) + ImplicitSourceTerm(var=T_e, coeff=G) - ImplicitSourceTerm(coeff=G, var=T_ph))
eq = eq0 & eq1
dt = 1e-18
steps = 7000
elapsed = 0.
vi = Viewer((T_e, T_ph), datamin=0., datamax=2e4)
for step in range(steps):
T_e.updateOld()
T_ph.updateOld()
vi.plot()
res = 1e100
dt *= 1.1
count = 0
while count < 10:
res = eq.sweep(dt=dt, underRelaxation=0.5)
print(t, res)
count += 1
print('elapsed:', t.value)
t.setValue(t + dt)
Regarding your questions.
I there any error in the fipy formalism of the equations?
Actually, no. Nothing wrong with the formalism, but better to use ImplicitSourceTerm.
What causes the error?
There are two source of instability in this system. The source terms inside the equation when written explicitly are unstable above a certain time step. Using an ImplcitSourceTerm removes this instablity. There is also some sort of instability in the coupling of the equations. I think that using under relaxation helps with that.
Is the error because of time step increase? If yes, how can I increase my time step?
Explained above.

In addition to #wd15's answer:
Your equations are extremely non-linear. You will likely benefit from Newton iterations to get decent convergence.
As #TimRoberts said, geometrically increasing the time step without bound is probably not a good idea.
I've recently posted a package called steppyngstounes that takes care of adapting timesteps. Although a standalone package, it's intended to work with FiPy. For example, you could change your solve loop to this:
from steppyngstounes import FixedStepper, PIDStepper
T_e.updateOld()
T_ph.updateOld()
for checkpoint in FixedStepper(start=0, stop=1e-10, size=1e-12):
for step in PIDStepper(start=checkpoint.begin,
stop=checkpoint.end,
size=dt):
res = 1e100
for sweep in range(10):
res = eq.sweep(dt=dt, underRelaxation=0.5)
print(t, sweep, res)
if step.succeeded(error=res / 1000):
T_e.updateOld()
T_ph.updateOld()
t.value = step.end
else:
T_e.value = T_e.old
T_ph.value = T_ph.old
print('elapsed:', t.value)
# the last step might have been smaller than possible,
# if it was near the end of the checkpoint range
dt = step.want
_ = checkpoint.succeeded()
vi.plot()
This code will update the viewer every 1e-12 time units, and adaptively make it's way between those checkpoints. There are other steppers in the package that would facilitate taking geometrically or exponentially increasing checkpoints, if that kept things more interesting.
You could probably get better overall performance by sweeping fewer times and letting the adapter take much smaller time steps in the beginning. I found that no time step was small enough to get the initial residual lower than 777.9. After the first couple of steps, the error metric could probably be much more aggressive, giving more accurate results.

Ways to Optimize Independent N-Body Simulations

I am relatively new to parallel computing and the Numba package. I am looking for optimization methods for my stupendously parallel N-body simulation. I've applied everything I know so far with Numpy arrays, JIT compliers, and multiprocessing. However, I'm still not getting the speed I desire (I've seen videos where their codes are MUCH faster still)
What I have currently is a rather simple python integrator using Runge-Kutta Integration and two equations of motion. I work with numerical integrators a lot in my field so I would definitely like to pick up a few more tricks from you guys.
I have posted my code below, but essentially, I have one main function called "Motion" which takes 2 initial conditions and integrate their motion for a set amount of time. I have JITTED this function and all the functions it called upon iteratively: "RK4", "ODE", "Electric Field". Lastly, I call the pool function from Multiprocessing to parallelize the "Motion" function and insert different initial conditions for each simulation it runs.
Again, I've implemented every types of optimization I'm aware of, however I'm still not very happy with its speed. I've posted my code below. If anyone can spot a piece of algorithm that could be further optimized, that would be extremely helpful and educational (for me at least)! Thank you for your time.
import numpy as np
import matplotlib.pyplot as plt
from numba import njit, prange
from time import time
from tqdm import tqdm
import multiprocessing as mp
from IPython.display import clear_output
from scipy import interpolate
"Electric Field Information"
A = np.float32(1.00E-04)
N_waves = np.int(19)
frequency = np.linspace(37.5,46.5,N_waves)*1e-3 #Set of frequencies used for Electric Field
m = np.int(20) #Azimuthal Wave Number
sigma = np.float32(0.5) #Gaussian Width of E wave in L
zeta = np.float32(1)
"Particle Information"
N_Particles = np.int(10000)
q = np.float32(-1) #Charge of electron
mass = np.float32(0.511e6) #Mass of Proton eV/c^2
FirstAdiabatic = np.float32(2000e10) #MeV/Gauss Adiabatic Invariant
"Runge-Kutta Paramters"
Total_Time = np.float32(10) #hours
Step_Size = np.float32(0.2) #second
Plot_Time = np.float32(60) #seconds
time_array = np.arange(0, Total_Time*3600+Step_Size, Step_Size) #Convert to seconds and Add End Point
N_points = len(time_array)
Skip_How_Many = int(Plot_Time/Step_Size) #Used to shorten our data set and save RAM
"Constants"
Beq = np.float64(31221.60592e-9) #nT
Re = np.float32(6371e3) #Meters
c = np.float32(2.998e8) #m/s
"Start Electric Field Code"
def wave_peak(omega): #Called once so no need to JIT or Optimize this
L_sample = np.linspace(1,10,100)
phidot = -3*FirstAdiabatic / (q* (L_sample*Re)**2 * np.sqrt(1+ (2*FirstAdiabatic*Beq/ (mass*L_sample**3)) ) )
phidot_to_L = interpolate.interp1d(phidot,L_sample, kind = 'cubic')
L0i = phidot_to_L(omega/m)
return L0i
omega = 2*np.pi*frequency
L0i_wave = wave_peak(omega)
Phi0i_wave = np.linspace(0,2*np.pi,N_waves)
np.random.shuffle(Phi0i_wave)
#njit(nogil= True)
def Electric_Field(t,r):
E0 = A*np.exp(-(r[0]-L0i_wave)**2 / (2*sigma**2))
Delta = np.arctan2( (r[0] * (r[0]-L0i_wave)/sigma**2 - 1), (2*np.pi*r[0]/zeta) )
Er = E0/m * np.sqrt( (2*np.pi*r[0]/zeta)**2 + (r[0]*(r[0]-L0i_wave)/sigma**2 -1)**2 ) * np.cos(m*r[1] - omega*t + Phi0i_wave + 2*np.pi*r[0]/zeta + Delta)
Ephi = E0*np.cos(m*r[1] - omega*t + Phi0i_wave + 2*np.pi*r[0]/zeta)
return np.sum(Er),np.sum(Ephi)
"End of Electric Field Code"
"Particle's ODE - Equation of Motion"
#njit(nogil= True)
def ODE(t,r):
Er, Ephi = Electric_Field(t,r) #Pull out the electric so we only call it once.
Ldot = Ephi * r[0]**3 / (Re*Beq)
Phidot = -Er * r[0]**2 / (Re*Beq) - 3* FirstAdiabatic / (q*r[0]**2*Re**2) * 1/np.sqrt(2*FirstAdiabatic*Beq/ (r[0]**3*mass) + 1)
return np.array([Ldot,Phidot])
#njit(nogil= True)
def RK4(t,r): #Standard Runge-Kutta Integration Algorthim
k1 = Step_Size*ODE(t,r)
k2 = Step_Size*ODE(t+Step_Size/2, r+k1/2)
k3 = Step_Size*ODE(t+Step_Size/2, r+k2)
k4 = Step_Size*ODE(t+Step_Size, r+k3)
return r + k1/6 + k2/3 + k3/3 + k4/6
#njit(nogil= True)
def Motion(L0,Phi0): #Insert Inital Conditions and it will loop through the RK4 integrator and output all its positions.
L_Array = np.zeros_like(time_array)
Phi_Array = np.zeros_like(time_array)
L_Array[0] = L0
Phi_Array[0] = Phi0
for i in range(1,N_points):
L_Array[i], Phi_Array[i] = RK4(time_array[i-1], np.array([ L_Array[i-1],Phi_Array[i-1] ]) )
return L_Array[::Skip_How_Many], Phi_Array[::Skip_How_Many]
#Skip_How_Many is used to take up less RAM space since we don't need that kind of percsion in our data
# Location = Motion(5,0)
# x = Location[0]*np.cos(Location[1])
# y = Location[0]*np.sin(Location[1])
# plt.plot(x,y,"o", markersize = 0.5)
# ts = time()
# Motion(5,0)
# print('Solo Time:', time() - ts)
"Getting my Arrays ready so I can index it"
Split = int(np.sqrt(N_Particles))
L0i = np.linspace(4.4,5.5,Split)
Phi0i = np.linspace(0,360,Split) / 180 * np.pi
L0_Grid = np.repeat(L0i,Split)
# ^Here I want to run a meshgrid of L0i and Phi0, so I repeat L0i using this function and mod (%) the index on the Phi Function
#Create Appending Array
results = []
def get_results(result): #Call back to this array from Multiprocessing to append the results it gives per run.
results.append(result)
clear_output()
print("Getting Results %0.2f" %(len(results)/N_Particles * 100), end='\r')
if __name__ == '__main__':
#Call In Multiprocessing
pool = mp.Pool(mp.cpu_count()) #Counting number of threads to start
ts = time() #Timing this process. Begins here
for ii in range(N_Particles): #Not too sure what this does, but it works - I assume it parallelizes this loop
pool.apply_async(Motion, args = (L0_Grid[ii],Phi0i[int(ii%Split)]), callback=get_results)
pool.close() #I'm not too sure what this does but everyone uses it, and it won't work without it
pool.join()
print('Time in MP parallel:', time() - ts) #Output Time

I think the main reason why your code is slow is because your Runge-Kutta method has fixed time steps. Fancy ODE solvers will select the biggest time step that allows a tolerable amount of error. One example is the LSODA ODE solver from ODEPACK.
Below I've re-written your code using NumbaLSODA. On my computer, it speeds up your code by about 200x.
import numpy as np
import matplotlib.pyplot as plt
from numba import njit, prange
from time import time
from tqdm import tqdm
import multiprocessing as mp
from scipy import interpolate
from NumbaLSODA import lsoda_sig, lsoda
from numba import cfunc
import numba as nb
"Electric Field Information"
A = np.float32(1.00E-04)
N_waves = np.int(19)
frequency = np.linspace(37.5,46.5,N_waves)*1e-3 #Set of frequencies used for Electric Field
m = np.int(20) #Azimuthal Wave Number
sigma = np.float32(0.5) #Gaussian Width of E wave in L
zeta = np.float32(1)
"Particle Information"
N_Particles = np.int(10000)
q = np.float32(-1) #Charge of electron
mass = np.float32(0.511e6) #Mass of Proton eV/c^2
FirstAdiabatic = np.float32(2000e10) #MeV/Gauss Adiabatic Invariant
"Runge-Kutta Paramters"
Total_Time = np.float32(10) #hours
Step_Size = np.float32(0.2) #second
Plot_Time = np.float32(60) #seconds
time_array = np.arange(0, Total_Time*3600+Step_Size, Step_Size) #Convert to seconds and Add End Point
N_points = len(time_array)
Skip_How_Many = int(Plot_Time/Step_Size) #Used to shorten our data set and save RAM
"Constants"
Beq = np.float64(31221.60592e-9) #nT
Re = np.float32(6371e3) #Meters
c = np.float32(2.998e8) #m/s
"Start Electric Field Code"
def wave_peak(omega): #Called once so no need to JIT or Optimize this
L_sample = np.linspace(1,10,100)
phidot = -3*FirstAdiabatic / (q* (L_sample*Re)**2 * np.sqrt(1+ (2*FirstAdiabatic*Beq/ (mass*L_sample**3)) ) )
phidot_to_L = interpolate.interp1d(phidot,L_sample, kind = 'cubic')
L0i = phidot_to_L(omega/m)
return L0i
omega = 2*np.pi*frequency
L0i_wave = wave_peak(omega)
Phi0i_wave = np.linspace(0,2*np.pi,N_waves)
np.random.shuffle(Phi0i_wave)
#njit
def Electric_Field(t,r):
E0 = A*np.exp(-(r[0]-L0i_wave)**2 / (2*sigma**2))
Delta = np.arctan2( (r[0] * (r[0]-L0i_wave)/sigma**2 - 1), (2*np.pi*r[0]/zeta) )
Er = E0/m * np.sqrt( (2*np.pi*r[0]/zeta)**2 + (r[0]*(r[0]-L0i_wave)/sigma**2 -1)**2 ) * np.cos(m*r[1] - omega*t + Phi0i_wave + 2*np.pi*r[0]/zeta + Delta)
Ephi = E0*np.cos(m*r[1] - omega*t + Phi0i_wave + 2*np.pi*r[0]/zeta)
return np.sum(Er),np.sum(Ephi)
"End of Electric Field Code"
"Particle's ODE - Equation of Motion"
#cfunc(lsoda_sig)
def ODE(t, r_, dr, p):
r = nb.carray(r_, (2,))
Er, Ephi = Electric_Field(t,r)
Ldot = Ephi * r[0]**3 / (Re*Beq)
Phidot = -Er * r[0]**2 / (Re*Beq) - 3* FirstAdiabatic / (q*r[0]**2*Re**2) * 1/np.sqrt(2*FirstAdiabatic*Beq/ (r[0]**3*mass) + 1)
dr[0] = Ldot
dr[1] = Phidot
funcptr = ODE.address
#njit
def Motion(L0,Phi0):
u0 = np.array([L0,Phi0],np.float64)
data = np.array([5.0])
usol, success = lsoda(funcptr, u0, time_array, data)
L_Array = usol[:,0]
Phi_Array = usol[:,1]
return L_Array[::Skip_How_Many], Phi_Array[::Skip_How_Many]
#Skip_How_Many is used to take up less RAM space since we don't need that kind of percsion in our data
Location = Motion(5,0)
x = Location[0]*np.cos(Location[1])
y = Location[0]*np.sin(Location[1])
plt.plot(x,y,"o", markersize = 0.5)
ts = time()
Motion(5,0)
print('Solo Time:', time() - ts)

Solving an Integral equation with uncertainties, by using fsolve and uncertainties packages in Python

I have some variables that are uncertain, these are
w_m = u.ufloat(0.1430, 0.0011)
z_rec = u.ufloat(1089.92, 0.25)
theta_srec = u.ufloat(0.0104110, 0.0000031)
r_srec = u.ufloat(144.43, 0.26)
and some constant values
c = 299792.458 # speed of light in [km/s]
N_eff = 3.046
w_r = 2.469 * 10**(-5) * (1 + (7/8)*(4/11)**(4/3) * N_eff)
My problem is I need to solve an integral defined by this functions
def D_zrec(z):
return (c/100) / sqrt(w_m * (1+z)**3 + w_r * (1+z)**4 + (h**2 - w_m - w_r))
This function is evaluated for dz but it also contains an unknown h that we need to find, with corresponding uncertainty. So I need to write a code that finds h.
Here is my full code
from numpy import sqrt, vectorize
from scipy.integrate import quad
import uncertainties as u
from uncertainties.umath import *
from scipy.optimize import fsolve
#### Important Parameters #####
c = 299792.458 # speed of light in [km/s]
N_eff = 3.046
w_r = 2.469 * 10**(-5) * (1 + (7/8)*(4/11)**(4/3) * N_eff)
w_m = u.ufloat(0.1430, 0.0011)
z_rec = u.ufloat(1089.92, 0.25)
theta_srec = u.ufloat(0.0104110, 0.0000031)
r_srec = u.ufloat(144.43, 0.26)
D_zrec_true = r_srec / theta_srec
#u.wrap
def D_zrec_finder(h, w_m, z_rec, D_zrec_true):
def D_zrec(z):
return (c/100) / sqrt(w_m * (1+z)**3 + w_r * (1+z)**4 + (h**2 - w_m - w_r))
result, error = quad(D_zrec, 0, z_rec)
return D_zrec_true - result
def h0_finder(w_m, z_rec, D_zrec_true):
vfunc = vectorize(D_zrec_finder)
sol = fsolve(vfunc, u.ufloat(0.6728, 0.01), args=(w_m, z_rec, D_zrec_true))[0]
return sol
print(h0_finder(w_m, z_rec, D_zrec_true))
So to summarize I have an integral named as D_zrec that is a function of z, but also contains an unknown number h that we need to find by using fsolve.
I have found 3 sites that might be useful for the coder. Please look at them if you want to help
https://kitchingroup.cheme.cmu.edu/blog/2013/03/07/Another-approach-to-error-propagation/
https://kitchingroup.cheme.cmu.edu/blog/2013/07/10/Uncertainty-in-an-integral-equation/
https://kitchingroup.cheme.cmu.edu/blog/2013/01/23/Solving-integral-equations-with-fsolve/
I have looked at them to write my code but no luck.
Thanks for the help

Determining lunar eclipse in skyfield

I am given a list of dates in UTC, all hours cast to 00:00.
I'd like to determine if a (lunar) eclipse occurred in a given day (ie past 24 hours)
Considering the python snippet
from sykfield.api import load
eph = load('de421.bsp')
def eclipticangle(t):
moon, earth = eph['moon'], eph['earth']
e = earth.at(t)
x, y, _ = e.observe(moon).apparent().ecliptic_latlon()
return x.degrees
I am assuming one is able to determine if an eclipse occurred within 24hrs of a time t by
Checking that the first angle is close enough to 180 (easy)
Checking if the second degree is close enough to 0 (not so essy?)
Now as far as the answer in the comment suggests it is not so trivial to solve the second problem simply by testing if the angle is close to 0.
Therefore, my question is
Can someone provide a function to determine if a lunar eclipse occurred on a given day t?
Edit. This question was edited to reflect the feedback from Brandon Rhodes left in the comments below.

I just went through section 11.2.3 of the Explanatory Supplement to the Astronomical Almanac and tried turning it into Skyfield Python code. Here is what I came up with:
import numpy as np
from skyfield.api import load
from skyfield.constants import ERAD
from skyfield.functions import angle_between, length_of
from skyfield.searchlib import find_maxima
eph = load('de421.bsp')
earth = eph['earth']
moon = eph['moon']
sun = eph['sun']
def f(t):
e = earth.at(t).position.au
s = sun.at(t).position.au
m = moon.at(t).position.au
return angle_between(s - e, m - e)
f.step_days = 5.0
ts = load.timescale()
start_time = ts.utc(2019, 1, 1)
end_time = ts.utc(2020, 1, 1)
t, y = find_maxima(start_time, end_time, f)
e = earth.at(t).position.m
m = moon.at(t).position.m
s = sun.at(t).position.m
solar_radius_m = 696340e3
moon_radius_m = 1.7371e6
pi_m = np.arcsin(ERAD / length_of(m - e))
pi_s = np.arcsin(ERAD / length_of(s - e))
s_s = np.arcsin(solar_radius_m / length_of(s - e))
pi_1 = 0.998340 * pi_m
sigma = angle_between(s - e, e - m)
s_m = np.arcsin(moon_radius_m / length_of(e - m))
penumbral = sigma < 1.02 * (pi_1 + pi_s + s_s) + s_m
partial = sigma < 1.02 * (pi_1 + pi_s - s_s) + s_m
total = sigma < 1.02 * (pi_1 + pi_s - s_s) - s_m
mask = penumbral | partial | total
t = t[mask]
penumbral = penumbral[mask]
partial = partial[mask]
total = total[mask]
print(t.utc_strftime())
print(0 + penumbral + partial + total)
It produces a vector of times at which lunar eclipses occurred, and then a rating of how total the eclipse is:
['2019-01-21 05:12:51 UTC', '2019-07-16 21:31:27 UTC']
[3 2]
Its eclipse times are within 3 seconds of the times given in the huge table of lunar ephemerides at NASA:
https://eclipse.gsfc.nasa.gov/5MCLE/5MKLEcatalog.txt