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solve two simultaneous equations: one contains a Python function

On the below map, I have two known points (A and B) with their coordinates (longitude, latitude). I need to derive the coordinates of a point C which is on the line, and is 100 kilometres away from A.
First I created a function to calculate the distances between two points in kilometres:
# pip install haversine
from haversine import haversine
def get_distance(lat_from,long_from,lat_to,long_to):
distance_in_km = haversine((lat_from,long_from),
(lat_to, long_to),
unit='km')
return distance_in_km
Then using the slope and the distance, the coordinates of point C should be the solution to the below equations:
# line segment AB and AC share the same slope, so
# (15.6-27.3)/(41.6-34.7) = (y-27.3)/(x-34.7)
# the distance between A and C is 100 km, so
# get_distance(y,x,27.3,34.7) = 100
Then I try to solve these two equations in Python:
from sympy import symbols, Eq, solve
slope = (15.6-27.3)/(41.6-34.7)
x, y = symbols('x y')
eq1 = Eq(y-slope*(x-34.7)-27.3)
eq2 = Eq(get_distance(y,x,34.7,27.3)-100)
solve((eq1,eq2), (x, y))
The error is TypeError: can't convert expression to float. I may understand the error, because the get_distance function is expecting inputs as floats, while my x and y in eq2 are sympy.core.symbol.Symbol.
I tried to add np.float(x), but the same error remains.
Is there a way to solve equations like these? Or do you have better ways to achieve what is needed?
Many thanks!
# there is a simple example of solving equations:
from sympy import symbols, Eq, solve
x, y = symbols('x y')
eq1 = Eq(2*x-y)
eq2 = Eq(x+2-y)
solve((eq1,eq2), (x, y))
# output: {x: 2, y: 4}

You can directly calculate that point. We can implement a python version of the intermediate calculation for lat long.
Be aware this calculations assume the earth is a sphere, and takes the curve into account, i.e. this is not a Euclidean approximation like your original post.
Say we have two (lat,long) points A and B;
import numpy as np
A = (52.234869, 4.961132)
B = (46.491267, 26.994655)
EARTH_RADIUS = 6371.009
We can than calculate the intermediate point fraction f by taking 100/distance-between-a-b-in-km
from sklearn.neighbors import DistanceMetric
dist = DistanceMetric.get_metric('haversine')
point_1 = np.array([A])
point_2 = np.array([B])
delta = dist.pairwise(np.radians(point_1), np.radians(point_2) )[0][0]
f = 100 / (delta * EARTH_RADIUS)
phi_1, lambda_1 = np.radians(point_1)[0]
phi_2, lambda_2 = np.radians(point_2)[0]
a = np.sin((1-f) * delta) / np.sin(delta)
b = np.sin( f * delta) / np.sin(delta)
x = a * np.cos(phi_1) * np.cos(lambda_1) + b * np.cos(phi_2) * np.cos(lambda_2)
y = a * np.cos(phi_1) * np.sin(lambda_1) + b * np.cos(phi_2) * np.sin(lambda_2)
z = a * np.sin(phi_1) + b * np.sin(phi_2)
phi_n = np.arctan2(z, np.sqrt(x**2 + y**2) )
lambda_n = np.arctan2(y,x)
The point C, going from A to B, with 100 km distance from A, is than
C = np.degrees( phi_n ), np.degrees(lambda_n)
In this case
(52.02172458025681, 6.384361456573444)

Specific heat fit using Debye+Einstein model in matplotlib

I am trying to find a fit to a specific heat data using gammaT+mDebye_model+(1-m)*Einstein model as given below.
Cel+ph(T ) = γ T + [αCDebye(T ) + (1 − α)CEinstein(T )]
where the Debye and Einstein models are given by eq. 3 and 4 in the attachment.
I have tried the following code in jupyter notebook following some examples on the web but i have no idea how can i combine these functions together to carry out the fit.
The data is linked https://www.dropbox.com/s/u0r2m3zwl8w77at/HC_ScPtBi.dat?dl=0
Column 1 is Temperature and Column 3 is Y data of interest.
Model is in https://www.dropbox.com/s/9452fq7eydajr5o/Debye.pdf?dl=0
Code is in https://www.dropbox.com/s/hk9b1t0agvt36zn/Untitled2.ipynb?dl=0
from matplotlib import pyplot
import numpy as np
from scipy import integrate
from scipy.optimize import curve_fit
from scipy.integrate import quad
data=np.genfromtxt('HC_ScPtBi.dat', skip_header=1)
R=8.314
n=3
M=1
T=data[10:290,0]
c=data[10:290,2]
def plot_data():
pyplot.scatter(T, c)
pyplot.xlabel('$T [K]$')
pyplot.ylabel('$C$')
plot_data()
def c_einstein(T, T_E):
x = T_E / T
return 3 *n*R*x**2 * np.exp(x) / (np.exp(x) - 1)**2
popt0, pcov0 = curve_fit(c_einstein, T, c, 250)
T_E = popt0[0]
delta_T_E = np.sqrt(pcov0[0, 0])
print(f"T_E = {T_E:.5} ± {delta_T_E:.3} K")
print(popt0)
plot_data()
#temps = np.linspace(10, T[-1], 100)
pyplot.plot(T, c_einstein(T, *popt0));
def integrand(y):
return y**4 * np.exp(y) / (np.exp(y) - 1)**2
#np.vectorize
def c_debye(T, T_D):
x = T / T_D
return 9 *n*R*x**3 * quad(integrand, 0, 1/x)[0]
popt1, pcov1 = curve_fit(c_debye, T, c, 150)
T_D = popt1[0]
delta_T_D = np.sqrt(pcov1[0, 0])
print(f"T_D = {T_D:.5} ± {delta_T_D:.3} K")
print(popt1)
plot_data()
pyplot.plot(T, c_einstein(T, *popt0), label='Einstein')
pyplot.plot(T, c_debye(T, *popt1), label='Debye')
pyplot.legend();

If it might be of any use, I obtained an excellent fit to a modified Weibull peak equation, with R-squared = 0.99999 and RMSE = 0.06878.
def Peak_WeibullPeak_Modified_model(x): # from zunzun.com
a = 6.4654735487019195E+01
b = 3.4517137038577323E+02
c = -1.5940608784806631E+00
d = 2.7331145870203617E+00
return = a * numpy.exp(-0.5 * numpy.power(numpy.log(x/b) / c, d))

You need to combine the Einstein and Debye equations into a single function, which should look something like this:
def func(T, alpha,gamma,T_e,T_d):
fn = lambda y: y**4 * np.exp(y) / (np.exp(y) - 1)**2
einst = (1-alpha)*3*n*R*T_e**2/T**2 * np.exp(T_e/T) / (np.exp(T_e/T) - 1)**2
debye_int = np.array([integrate.quad(fn, 0, T_d/t)[0] for t in T])
debye = alpha*9*n*R*T**3/T_d**3*debye_int
return einst+debye+gamma*T
You can then use that function in the curve fitting
coefs = curve_fit(func, T, c)[0]
plt.plot(T, func(T, *coefs))

Python integration of large array of coefficient

I'm trying to optimize simple integration in python which looks something like
from scipy import integrate
import numpy as np
from scipy.special import kv
import time
#Example function
def integrand(x, a, b, c):
return a * (x ** (-b)) * (np.sqrt(x ** (c) + 1) - 1)
#Real Function that I want to calculate
def Bes(xx):
return integrate.quad(lambda x: kv(5./3.,x), xx,np.inf)
def F(x,a,b,c,d,e,f):
zx = 1/((x**2.+1)*a)
feq = e*x**(f)
if (x>c):
feq *= c/x * np.exp(-(x/d)**2.)
return b*Bes(zx)*feq*x**2.
start = time.time()
array_length = 10
a = np.random.rand(array_length)+3.
b = np.random.rand(array_length)+1.
c = np.random.rand(array_length)
d = (np.random.rand(array_length)+1)*100.
e = np.random.rand(array_length)*100.
f = np.random.rand(array_length)
inte = np.array([])
for i in range(array_length):
result = integrate.quad(lambda x: F(x, a[i], b[i], c[i],d[i],e[i],f[i]),0.01,100000.)
inte = np.append(inte,result[0])
print("For array length = %i" % array_length)
print("Time = %.2f [sec]" %(time.time()-start))
But the problems that I'm facing are
a, b, c are array with length > 10^7 (same length)
integration range of x starts at 0.01 and extends to infinite
Integration at the small x (like [0.01, 1]) is very important and needs small step.
I want to integrate this function on each coefficient value and returns the entire array of integration as the result (length ~ 10^7), efficiently.
What kind of tools should I use?
(+) I just changed my code from simple example to actual integration form that I need to solve. Sorry for making confusion.

I suspected that this integral would converge for certain values of b and c, so I tried to evaluate this using Sympy:
import sympy
sympy.init_printing()
a, b, c = sympy.symbols('a, b, c', positive=True)
x = sympy.Symbol('x', positive=True)
sympy.integrate(a*(x**(-b))*(sympy.sqrt(x**c+1)-1), (x, 0, sympy.oo))
This means that you should be able to obtain the correct results with this code as long as your coefficients pass the check function.
from numpy import sqrt, pi
from scipy.special import gamma
def check(a, b, c):
assert (-(-b + 1)/c < 1)
assert (1/2 - (-b + 1)/c > 1)
assert (1 - (-b + 1)/c > 1)
def result(a, b, c):
return a*gamma(-b/c + 1 + 1/c)*gamma(b/c - 1/2 - 1/c)/(2*sqrt(pi)*(b - 1))

Generating random numbers a, b, c such that a^2 + b^2 + c^2 = 1

To do some simulations in Python, I'm trying to generate numbers a,b,c such that a^2 + b^2 + c^2 = 1. I think generating some a between 0 and 1, then some b between 0 and sqrt(1 - a^2), and then c = sqrt(1 - a^2 - b^2) would work.
Floating point values are fine, the sum of squares should be close to 1. I want to keep generating them for some iterations.
Being new to Python, I'm not really sure how to do this. Negatives are allowed.
Edit: Thanks a lot for the answers!

According to this answer at stats.stackexchange.com, you should use normally distributed values to get uniformly distributed values on a sphere. That would mean, you could do:
import numpy as np
abc = np.random.normal(size=3)
a,b,c = abc/np.sqrt(sum(abc**2))

Just in case your interested in the probability densities I decided to do a comparison between the different approaches:
import numpy as np
import random
import math
def MSeifert():
a = 1
b = 1
while a**2 + b**2 > 1: # discard any a and b whose sum of squares already exceeds 1
a = random.random()
b = random.random()
c = math.sqrt(1 - a**2 - b**2) # fixed c
return a, b, c
def VBB():
x = np.random.uniform(0,1,3) # random numbers in [0, 1)
x /= np.sqrt(x[0] ** 2 + x[1] ** 2 + x[2] ** 2)
return x[0], x[1], x[2]
def user3684792():
theta = random.uniform(0, 0.5*np.pi)
phi = random.uniform(0, 0.5*np.pi)
return np.sin(theta)* np.cos(phi), np.sin(theta)*np.sin(phi), np.cos(theta)
def JohanL():
abc = np.random.normal(size=3)
a,b,c = abc/np.sqrt(sum(abc**2))
return a, b, c
def SeverinPappadeux():
cos_th = 2.0*random.uniform(0, 1.0) - 1.0
sin_th = math.sqrt(1.0 - cos_th*cos_th)
phi = random.uniform(0, 2.0*math.pi)
return sin_th * math.cos(phi), sin_th * math.sin(phi), cos_th
And plotting the distributions:
%matplotlib notebook
import matplotlib.pyplot as plt
f, axes = plt.subplots(3, 4)
for func_idx, func in enumerate([MSeifert, JohanL, user3684792, VBB]):
axes[0, func_idx].set_title(str(func.__name__))
res = [func() for _ in range(50000)]
for idx in range(3):
axes[idx, func_idx].hist([i[idx] for i in res], bins='auto')
axes[0, 0].set_ylabel('a')
axes[1, 0].set_ylabel('b')
axes[2, 0].set_ylabel('c')
plt.tight_layout()
With the result:
Explanation: The rows show the distributions for a, b and c respectively while the columns show the histograms (distributions) of the different approaches.
The only approaches that give a uniformly random distribution in the range (-1, 1) are JohanLs and Severin Pappadeux's approach. All other approaches have some features like spikes or a functional behavior in the range [0, 1). Note that these two solution currently gives values between -1 and 1 while all other approaches give values between 0 and 1.

I think it is actually a cool problem, and a nice way to do this is to just use spherical polar coordinates and generate the angles at random.
import random
import numpy as np
def random_pt():
theta = random.uniform(0, 0.5*np.pi)
phi = random.uniform(0, 0.5*np.pi)
return np.sin(theta)* np.cos(phi), np.sin(theta)*np.sin(phi), np.cos(theta)

You could do it like this:
import random
import math
def three_random_numbers_adding_to_one():
a = 1
b = 1
while a**2 + b**2 > 1: # discard any a and b whose sum of squares already exceeds 1
a = random.random()
b = random.random()
c = math.sqrt(1 - a**2 - b**2) # fixed c
return a, b, c
a, b, c = three_random_numbers_adding_to_one()
print(a**2 + b**2 + c**2)
However floats have only limited precision so these won't add to exactly 1, just approximately.
You may need to check if the numbers generated with this function are "random enough". It could be that this setup biases the "randomness".

The "right" answer depends on whether you are looking for a uniform random distribution in space, or on the surface of a sphere, or something else. If you are looking for points on the surface of a sphere, you still have to worry about the cos(theta) factor which will cause points to appear "bunched up" near the poles of the sphere. Since exact nature is not clear from your question, here is a "totally random" distribution that should work:
x = np.random.uniform(0,1,3) # random numbers in [0, 1)
x /= np.sqrt(x[0] ** 2 + x[1] ** 2 + x[2] ** 2)
Another advantage here is that since we are using numpy arrays, you can quickly scale to large sets of points too, by using x = np.random.uniform(0, 1, (3, n)) for any n.

Time to add another solution, heh...
This time it is truly uniform on the unit sphere point picking - check http://mathworld.wolfram.com/SpherePointPicking.html for details
import math
import random
def random_pt():
cos_th = 2.0*random.uniform(0, 1.0) - 1.0
sin_th = math.sqrt(1.0 - cos_th*cos_th)
phi = random.uniform(0, 2.0*math.pi)
return sin_th * math.cos(phi), sin_th * math.sin(phi), cos_th
for k in range(0, 100):
a, b, c = random_pt()
print("{0} {1} {2} {3}".format(a, b, c, a*a + b*b + c*c))

Python double integral taking too long to compute

I am trying to compute the fresnel integral over a grid of coordinates using dblquad. But its taking very long and finally it's not giving any result.
Below is my code. In this code I integrated only over a 10 x 10 grid but I need to integrate at least over a 500 x 500 grid.
import time
st = time.time()
import pylab
import scipy.integrate as inte
import numpy as np
print 'imhere 0'
def sinIntegrand(y,x, X , Y):
a = 0.0001
R = 2e-3
z = 10e-3
Lambda = 0.5e-6
alpha = 0.01
k = np.pi * 2 / Lambda
return np.cos(k * (((x-R)**2)*a + (R-(x**2 + y**2)) * np.tan(np.radians(alpha)) + ((x - X)**2 + (y - Y)**2) / (2 * z)))
print 'im here 1'
def cosIntegrand(y,x,X,Y):
a = 0.0001
R = 2e-3
z = 10e-3
Lambda = 0.5e-6
alpha = 0.01
k = np.pi * 2 / Lambda
return np.sin(k * (((x-R)**2)*a + (R-(x**2 + y**2)) * np.tan(np.radians(alpha)) + ((x - X)**2 + (y - Y)**2) / (2 * z)))
def y1(x,R = 2e-3):
return (R**2 - x**2)**0.5
def y2(x, R = 2e-3):
return -1*(R**2 - x**2)**0.5
points = np.linspace(-1e-3,1e-3,10)
points2 = np.linspace(1e-3,-1e-3,10)
yv,xv = np.meshgrid(points , points2)
#def integrate_on_grid(func, lo, hi,y1,y2):
# """Returns a callable that can be evaluated on a grid."""
# return np.vectorize(lambda n,m: dblquad(func, lo, hi,y1,y2,(n,m))[0])
#
#intensity = abs(integrate_on_grid(sinIntegrand,-1e-3 ,1e-3,y1, y2)(yv,xv))**2 + abs(integrate_on_grid(cosIntegrand,-1e-3 ,1e-3,y1, y2)(yv,xv))**2
Intensity = []
print 'im here2'
for i in points:
row = []
for j in points2:
print 'im here'
intensity = abs(inte.dblquad(sinIntegrand,-1e-3 ,1e-3,y1, y2,(i,j))[0])**2 + abs(inte.dblquad(cosIntegrand,-1e-3 ,1e-3,y1, y2,(i,j))[0])**2
row.append(intensity)
Intensity.append(row)
Intensity = np.asarray(Intensity)
pylab.imshow(Intensity,cmap = 'gray')
pylab.show()
print str(time.time() - st)
I would really appreciate if you could tell any better way of doing this.

Using a scipy.integrate.dblquad to calculate every pixel of your image is going to be slow in any case.
You should try rewriting your mathematical problem so you can use some classical function in scipy.special instead. For instance, scipy.special.fresnel might work, although it is 1D and your problem seems to be in 2D. Otherwise, that there is a relationship between the Fresnel integral and the incomplete Gamma function (scipy.special.gammainc), if that helps.
If none of this work, as a last resort you can spend time optimizing your code and adapting it to Cython. This it will probably give a speed up of a factor of 10 to 100 (see this answer). Though this wouldn't be sufficient to go from a grid 10x10 to a grid 500x500.