import sympy
import numpy
from sympy import ordered, Matrix
x1,x2=sympy.symbols('x1 x2')
f=x1**2-x1*x2-4*x1+x2**2-x2
X0=numpy.array([[1],[1]])
v = list(ordered(f.free_symbols))
gradient = lambda f, v: Matrix([f]).jacobian(v)
gradf=sympy.transpose(gradient(f, v))
gradfx0=(gradf.subs([(x1, X0[0]), (x2, X0[1])]))
print(gradfx0)
I want to calculate the gradient of two variable function in a point in python. I define the function and I find the gradient vector of function (grad). Now when I try to substituting X0 to grad, the result is
Matrix([[2*x1 - x2 - 4], [-x1 + 2*x2 - 1]]).
I want the result should be
Matrix([[-3], [0]]).
How to substituting a point to sympy array?
Using .subs and derive_by_array:
from sympy import symbols
from sympy.tensor.array import derive_by_array
x1, x2 = symbols('x1 x2')
f = x1 ** 2 - x1 * x2 + x2 ** 2 - 4 * x1 - x2
grad = derive_by_array(f, (x1, x2))
# = [2*x1 - x2 - 4, -x1 + 2*x2 - 1]
gradx0 = grad.subs({x1: 1, x2: 1})
# = [-3, 0]
If you want to call your point x0 first, then use variable x0:
x0 = (1, 1)
gradx0 = grad.subs(zip((x1,x2), x0))
# = [-3, 0]
Related
I am trying to find a common tangent to two curves using python but I am not able to solve it.
The equations to the two curves are complicated that involve logarithms.
Is there a way in python to compute the x coordinates of a tangent that is common to both the curves in general. If I have 2 curves f(x) and g(x), I want to find the x-coordinates x1 and x2 on a common tangent where x1 lies on f(x) and x2 on g(x). I am trying f'(x1) = g'(x2) and f'(x1) = f(x1) - f(x2) / (x1 - x2) to get x1 and x2 but I am not able to get values using nonlinsolve as the equations are too complicated.
I want to just find x-coordinates of the common tangent
Can anyone suggest a better way?
import numpy as np
import sympy
from sympy import *
from matplotlib import pyplot as plt
x = symbols('x')
a, b, c, d, e, f = -99322.50019502985, -86864.87072433547, -96876.05627516498, -89703.35055202093, -3390.863799999999, -20942.518
def func(x):
y1_1 = a - a*x + b*x
y1_2 = c - c*x + d*x
c1 = (1 - x) ** (1 - x)
c2 = (x ** x)
y2 = 12471 * (sympy.log((c1*c2)))
y3 = 2*f*x**3 - x**2*(e + 3*f) + x*(e + f)
eqn1 = y1_1 + y2 + y3
eqn2 = y1_2 + y2 + y3
return eqn1, eqn2
val = np.linspace(0, 1)
f1 = sympy.lambdify(x, func(x)[0])(val)
f2 = sympy.lambdify(x, func(x)[1])(val)
plt.plot(val, f1)
plt.plot(val, f2)
plt.show()
I am trying this
x1, x2 = sympy.symbols('x1 x2')
fun1 = func(x1)[0]
fun2 = func(x2)[0]
diff1 = diff(fun1,x1)
diff2 = diff(fun2,x2)
eq1 = diff1 - diff2
eq2 = diff1 - ((fun1 - fun2) / (x1 - x2))
sol = nonlinsolve([eq1, eq2], [x1, x2])
the first thing that needs to be done is to reduce the formulas
for example the first formula is actually this:
formula = x*(1 - x)*(17551.6542 - 41885.036*x) + x*(1 - x)*(41885.036*x - 24333.3818) + 12457.6294706944*x + log((x/(1 - x))**(12000*x)*(1 - x)**12000) - 99322.5001950298
formula = (x-x^2)*(17551.6542 - 41885.036*x) + (x-x^2)*(41885.036*x - 24333.3818) + 12457.6294706944*x + log((x/(1 - x))**(12000*x)*(1 - x)**12000) - 99322.5001950298
# constants
a = 41885.036
b = 17551.6542
c = 24333.3818
d = 12457.6294706944
e = 99322.5001950298
f = 12000
formula = (x-x^2)*(b - a*x) + (x-x^2)*(a*x - c) + d*x + log((x/(1 - x))**(f*x)*(1 - x)**f) - e
formula = (ax^3 -bx^2 + bx - ax^2) + (x-x^2)*(a*x - c) + d*x + log((x/(1 - x))**(f*x)*(1 - x)**f) - e
formula = ax^3 -bx^2 + bx - ax^2 -ax^3 + ax^2 + cx^2 -cx + d*x + log((x/(1 - x))**(f*x)*(1 - x)**f) - e
# collect x terms by power (note how the x^3 tern drops out, so its easier).
formula = (c-b)*x^2 + (b-c+d)*x + log((x/(1 - x))**(f*x)*(1 - x)**f) - e
which is much cleaner and is a quadratic with a log term.
i expect that you can do some work on the log term too, but this is an excercise for the original poster.
likewise the second formula can be reduced in the same way, which is again an excercise for the original poster.
From this, both equations need to be differentiated with respect to x to find the tangent. Then set both formulas to be equal to each other (for a common tangent).
This would completely solve the question.
I actually wonder if this is a python question at all or actually a pure maths question.....
The important point to note is that, since the derivatives are monotonic, for any value of derivative of fun1, there is a solution for fun2. This can be easily seen if you plot both derivatives.
Thus, we want a function that, given an x1, returns an x2 that matches it. I'll use numerical solution because the system is too cumbersome for numerical solution.
import scipy.optimize
def find_equal_value(f1, f2, x, x1):
goal = f1.subs(x, x1)
to_solve = sympy.lambdify(x, (f2 - goal)**2) # Quadratic functions tend to be better behaved, and the result is the same
sol = scipy.optimize.fmin(func=to_solve, x0=x1, ftol=1e-8, disp=False) # The value for f1 is a good starting guess
return sol[0]
I used fmin as the solver above because it worked and I knew how to use it by heart. Maybe root_scalar can give better results.
Using the function above, let's get some pairs (x1, x2) where the derivatives are equal:
df1 = sympy.diff(func(x)[0])
df2 = sympy.diff(func(x)[1])
x1 = 0.25236537 # Close to the zero derivative
x2 = find_equal_value(df1, df2, x, x1)
print(f'Derivative of f1 in x1: {df1.subs(x, x1)}')
print(f'Derivative of f2 in x2: {df2.subs(x, x2)}')
print(f'Error: {df1.subs(x, x1) - df2.subs(x, x2)}')
This results is:
Derivative of f1 in x1: 0.0000768765858083498
Derivative of f2 in x2: 0.0000681969431752805
Error: 0.00000867964263306931
If you want a x2 for several x1s (beware that in some cases the solver hits a value where the logs are invalid. Always check your result for validity):
x1s = np.linspace(0.2, 0.8, 50)
x2s = [find_equal_value(df1, df2, x, x1) for x1 in x1s]
plt.plot(x1s, x2s); plt.grid(); plt.show()
I have tried adapting euler method code, and am using euler to calculate the first value so there are two for me to use in verlet but when i plot the graph i just get two perpendicular straight lines.
this is the code:
for t in t_array:
if t == 0:
x0 = x
x1 = x0
a = -k * x1 / m
x2 = x1 + dt * v
v = v + dt * a
x_list.append(x2)
v_list.append(v)
else:
x2 = x1
x1 = x0
x2 = 2*x1 - x0 + dt**2*a
v = (1/dt)*(x2 - x1)
x_list.append(x2)
v_list.append(v)
and then i plot a graph using matplotlib.
I am trying to find the gradient of the function
f(x) = w1 * x1^2 + w2 * x2
where x is a vector coordinate (x1,x2).
def gradient(w1, w2, x):
x= (x1,x2)
gradx1=2*w1*x1 + w2 * x2
gradx2= w2 + w1 * x1^2
return (gradx1, gradx2)
My code is coming up with a nameError, saying x1 is not defined when calling the function:
gradient(5, 6, (10,10))
First things first:
x1, x2 = x # unpack your coord tuple
And secondly:
gradx2= w2 + w1 * x1 ** 2 # or gradx2= w2 + w1 * x1 * x1
in python ^ is bitwise XOR. Exponentiation is **.
x is a tuple which you need to unpack like so:
x1, x2 = x
Rather than:
x = (x1, x2)
I have two arrays as follows:
X1 = np.array([[x11, x12, x13 ... x1n],[y11, y12, y13, ... , y1n]])
X2 = np.array([[x21, x22, x23 ... x2n],[y21, y22, y23, ... , y2n]])
I would like to basically conceptualize these as piecewise linear functions and come up with an intersection point intercept:
intercept = (x_int, y_int)
Every search I do regarding array intersection gives entirely unrelated results, since the intersection of two arrays also has the meaning of finding elements common to both arrays (rather than an intersection point).
I also found this interesting post, but it seems too complex for my application. If I had to implement this, I think I could since it would involve repeated calculations of line equations and intersections of points between line equations. However I'm first trying to check if some robust implementation already exists in a well-tested library, since my poor attempt might take hours/days to achieve and then not necessarily be applicable for any dataset.
Has this already been implemented in python?
Step 1: work with the union of x1 and x2.
Step 2: linearly interpolate to find y1 and y2 for each point in the union.
Step 3: find where y1 - y2 changes sign.
Step 4: solve the linear equations for the intersection.
import numpy as np
def intersect_piecewise(X1, X2):
x = np.union1d(X1[0], X2[0])
y1 = np.interp(x, X1[0], X1[1])
y2 = np.interp(x, X2[0], X2[1])
dy = y1 - y2
ind = (dy[:-1] * dy[1:] < 0).nonzero()[0]
x1, x2 = x[ind], x[ind+1]
dy1, dy2 = dy[ind], dy[ind+1]
y11, y12 = y1[ind], y1[ind+1]
y21, y22 = y2[ind], y2[ind+1]
x_int = x1 - (x2 - x1) * dy1 / (dy2 - dy1)
y_int = y11 + (y12 - y11) * (x_int - x1) / (x2 - x1)
return x_int, y_int
the equations you need to solve are
(x_int - x1) / (x2 - x1) = (0 - dy1) / (dy2 - dy1)
= (y_int - y11) / (y12 - y11) = (y_int - y21) / (y22 - y21)
Edit: Let's try it out
import matplotlib.pyplot as plt
x = np.linspace(-2, 2, 17)
X1 = np.stack((x[::2], x[::2]**2))
X2 = np.stack((x[1::2], 4 - x[1::2]**2))
x_int, y_int = intersect_piecewise(X1, X2)
plt.plot(X1[0], X1[1], 'bo-', X2[0], X2[1], 'bo-', x_int, y_int, 'rs')
I am trying to solve for C in the following equation
I can do this with sympy for an enumrated number of x's, e.g x0, x2, ..., x4 but cannot seem to figure out how to do this for i=0 to t. E.g. for a limited number
from sympy import summation, symbols, solve
x0, x1, x2, x3, x4, alpha, C = symbols('x0, x1, x2, x3, x4, alpha, C')
e1 = ((x0 + alpha * x1 + alpha**(2) * x2 + alpha**(3) * x3 + alpha**(4) * x4)
/ (1 + alpha + alpha**(2) + alpha**(3) + alpha**(4)))
e2 = (x3 + alpha * x4) / (1 + alpha)
rhs = (x0 + alpha * x1 + alpha**(2) * x2) / (1 + alpha + alpha**(2))
soln_C = solve(e1 - C*e2 - rhs, C)
Any insight would be much appreciated.
Thanks to #bryans for pointing me in the direction of Sum. Elaborating on his comment, here is one solution that seems to work. As I am fairly new to sympy if anyone has a more concise approach please share.
from sympy import summation, symbols, solve, Function, Sum
alpha, C, t, i = symbols('alpha, C, t, i')
x = Function('x')
s1 = Sum(alpha**i * x(t-i), (i, 0, t)) / Sum(alpha**i, (i, 0, t))
s2 = Sum(alpha**i * x(t-3-i), (i, 0, t-3)) / Sum(alpha**i, (i, 0, t-3))
rhs = (x(0) + alpha * x(1) + alpha**(2) * x(2)) / (1 + alpha + alpha**(2))
soln_C = solve(s1 - C*s2 - rhs, C)
I'm not sure if this can be catalogued as more "concise", but it could be useful too, when you know the upper limit of the summatory. Let's suppose that we want to evaluate this expression:
We can express it, and solve it, in sympy as follows:
from sympy import init_session
init_session(use_latex=True)
n = 4
As = symbols('A_1:' + str(n+1))
x = symbols('x')
exp = 0
for i in range(n):
exp += As[i]/(1+x)**(i+1)
Ec = Eq(exp,0)
sol = solve(Ec,x)
#print(sol)
#sol #Or, if you're working on jupyter...