Suppose I have an eq1 such that
from sympy import symbols, solve, plot, Eq, diff
a, b, X, Y, U = symbols('a b X Y U')
eq1 = Eq(U, X**a*Y**b)
$U=(X^a)(Y^b)$
but When I run diff(eq1 , X)
the differential does not evaluate I merele just get the DU/DX symbol but not evaluated
I know I could defined the function as
U = X**a * Y**b
and easily compute diff(U)
but printing the U expression will not look nice.
I am surprised that you even get what you got. In general, algebraic operations on Eq are not supported.
>>> from sympy import Derivative
>>> ediff=lambda e, *x: e.func(Derivative(e.lhs,*x), e.rhs.diff(*x))
>>> ediff(eq1, X)
Eq(Derivative(U, X), X**a*Y**b*a/X)
I want to work with generic functions as long as possible, and only substitute functions at the end.
I'd like to define a function as the derivative of another one, define a generic expression with the function and its derivative, and substitute the function at the end.
Right now my attempts is as follows, but I get the error 'Derivative' object is not callable:
from sympy import Function
x, y, z = symbols('x y z')
f = Function('f')
df = f(x).diff(x) # <<< I'd like this to be a function of dummy variable x
expr = f(x) * df(z) + df(y) + df(0) # df is unfortunately not callable
# At the end, substitute with example function
expr.replace(f, Lambda(X, cos(X))) # should return: -cos(x)*sin(z) - sin(y) - sin(0)
I think I got it to work with integrals as follows:
I= Lambda( x, integrate( f(y), (y, 0, x))) but that won't work for derivatives.
If that helps, I'm fine restricting myself to functions of a single variable for now.
As a bonus, I'd like to get this to work with any combination (products, derivatives, integrals) of the original function.
It's pretty disappointing that f.diff(x) doesn't work, as you say. Maybe someone will create support it sometime in the future. In the mean time, there are 2 ways to go about it: either substitute x for your y, z, ... OR lambdify df.
I think the first option will work more consistently in the long run (for example, if you decide to extend to multivariate calculus). But the expr in second option is far more natural.
Using substitution:
from sympy import *
x, y, z = symbols('x y z')
X = Symbol('X')
f = Function('f')
df = f(x).diff(x)
expr = f(x) * df.subs(x, z) + df.subs(x, y) + df.subs(x, 0)
print(expr.replace(f, Lambda(X, cos(X))).doit())
Lambdifying df:
from sympy import *
x, y, z = symbols('x y z')
X = Symbol('X')
f = Function('f')
df = lambda t: f(t).diff(t) if isinstance(t, Symbol) else f(X).diff(X).subs(X, t)
expr = f(x) * df(z) + df(y) + df(0)
print(expr.replace(f, Lambda(X, cos(X))).doit())
Both give the desired output.
Let's suppose that I have two transcendental functions f(x, y) = 0 and g(a, b) = 0.
a and b depend on y, so if I could solve the first equation analytically for y, y = f(x), I could have the second function depending only on x and thus solving it numerically.
I prefer to use python, but if matlab is able to handle this is ok for me.
Is there a way to solve analytically trascendent functions for a variable with python/matlab? Taylor is fine too, as long as I can choose the order of approximation.
I tried running this through Sympy like so:
import sympy
j, k, m, x, y = sympy.symbols("j k m x y")
eq = sympy.Eq(k * sympy.tan(y) + j * sympy.tan(sympy.asin(sympy.sin(y) / x)), m)
eq.simplify()
which turned your equation into
Eq(m, j*sin(y)/(x*sqrt(1 - sin(y)**2/x**2)) + k*tan(y))
which, after a bit more poking, gives us
k * tan(y) + j * sin(y) / sqrt(x**2 - sin(y)**2) == m
We can find an expression for x(y) like
sympy.solve(eq, x)
which returns
[-sqrt(j**2*sin(y)**2/(k*tan(y) - m)**2 + sin(y)**2),
sqrt(j**2*sin(y)**2/(k*tan(y) - m)**2 + sin(y)**2)]
but an analytic solution for y(x) fails.
I have 4 non-linear equations with three unknowns X, Y, and Z that I want to solve for. The equations are of the form:
F(m) = X^2 + a(m)Y^2 + b(m)XYcosZ + c(m)XYsinZ
...where a, b and c are constants which are dependent on each value of F in the four equations.
What is the best way to go about solving this?
There are two ways to do this.
Use a non-linear solver
Linearize the problem and solve it in the least-squares sense
Setup
So, as I understand your question, you know F, a, b, and c at 4 different points, and you want to invert for the model parameters X, Y, and Z. We have 3 unknowns and 4 observed data points, so the problem is overdetermined. Therefore, we'll be solving in the least-squares sense.
It's more common to use the opposite terminology in this case, so let's flip your equation around. Instead of:
F_i = X^2 + a_i Y^2 + b_i X Y cosZ + c_i X Y sinZ
Let's write:
F_i = a^2 + X_i b^2 + Y_i a b cos(c) + Z_i a b sin(c)
Where we know F, X, Y, and Z at 4 different points (e.g. F_0, F_1, ... F_i).
We're just changing the names of the variables, not the equation itself. (This is more for my ease of thinking than anything else.)
Linear Solution
It's actually possible to linearize this equation. You can easily solve for a^2, b^2, a b cos(c), and a b sin(c). To make this a bit easier, let's relabel things yet again:
d = a^2
e = b^2
f = a b cos(c)
g = a b sin(c)
Now the equation is a lot simpler: F_i = d + e X_i + f Y_i + g Z_i. It's easy to do a least-squares linear inversion for d, e, f, and g. We can then get a, b, and c from:
a = sqrt(d)
b = sqrt(e)
c = arctan(g/f)
Okay, let's write this up in matrix form. We're going to translate 4 observations of (the code we'll write will take any number of observations, but let's keep it concrete at the moment):
F_i = d + e X_i + f Y_i + g Z_i
Into:
|F_0| |1, X_0, Y_0, Z_0| |d|
|F_1| = |1, X_1, Y_1, Z_1| * |e|
|F_2| |1, X_2, Y_2, Z_2| |f|
|F_3| |1, X_3, Y_3, Z_3| |g|
Or: F = G * m (I'm a geophysist, so we use G for "Green's Functions" and m for "Model Parameters". Usually we'd use d for "data" instead of F, as well.)
In python, this would translate to:
def invert(f, x, y, z):
G = np.vstack([np.ones_like(x), x, y, z]).T
m, _, _, _ = np.linalg.lstsq(G, f)
d, e, f, g = m
a = np.sqrt(d)
b = np.sqrt(e)
c = np.arctan2(g, f) # Note that `c` will be in radians, not degrees
return a, b, c
Non-linear Solution
You could also solve this using scipy.optimize, as #Joe suggested. The most accessible function in scipy.optimize is scipy.optimize.curve_fit which uses a Levenberg-Marquardt method by default.
Levenberg-Marquardt is a "hill climbing" algorithm (well, it goes downhill, in this case, but the term is used anyway). In a sense, you make an initial guess of the model parameters (all ones, by default in scipy.optimize) and follow the slope of observed - predicted in your parameter space downhill to the bottom.
Caveat: Picking the right non-linear inversion method, initial guess, and tuning the parameters of the method is very much a "dark art". You only learn it by doing it, and there are a lot of situations where things won't work properly. Levenberg-Marquardt is a good general method if your parameter space is fairly smooth (this one should be). There are a lot of others (including genetic algorithms, neural nets, etc in addition to more common methods like simulated annealing) that are better in other situations. I'm not going to delve into that part here.
There is one common gotcha that some optimization toolkits try to correct for that scipy.optimize doesn't try to handle. If your model parameters have different magnitudes (e.g. a=1, b=1000, c=1e-8), you'll need to rescale things so that they're similar in magnitude. Otherwise scipy.optimize's "hill climbing" algorithms (like LM) won't accurately calculate the estimate the local gradient, and will give wildly inaccurate results. For now, I'm assuming that a, b, and c have relatively similar magnitudes. Also, be aware that essentially all non-linear methods require you to make an initial guess, and are sensitive to that guess. I'm leaving it out below (just pass it in as the p0 kwarg to curve_fit) because the default a, b, c = 1, 1, 1 is a fairly accurate guess for a, b, c = 3, 2, 1.
With the caveats out of the way, curve_fit expects to be passed a function, a set of points where the observations were made (as a single ndim x npoints array), and the observed values.
So, if we write the function like this:
def func(x, y, z, a, b, c):
f = (a**2
+ x * b**2
+ y * a * b * np.cos(c)
+ z * a * b * np.sin(c))
return f
We'll need to wrap it to accept slightly different arguments before passing it to curve_fit.
In a nutshell:
def nonlinear_invert(f, x, y, z):
def wrapped_func(observation_points, a, b, c):
x, y, z = observation_points
return func(x, y, z, a, b, c)
xdata = np.vstack([x, y, z])
model, cov = opt.curve_fit(wrapped_func, xdata, f)
return model
Stand-alone Example of the two methods:
To give you a full implementation, here's an example that
generates randomly distributed points to evaluate the function on,
evaluates the function on those points (using set model parameters),
adds noise to the results,
and then inverts for the model parameters using both the linear and non-linear methods described above.
import numpy as np
import scipy.optimize as opt
def main():
nobservations = 4
a, b, c = 3.0, 2.0, 1.0
f, x, y, z = generate_data(nobservations, a, b, c)
print 'Linear results (should be {}, {}, {}):'.format(a, b, c)
print linear_invert(f, x, y, z)
print 'Non-linear results (should be {}, {}, {}):'.format(a, b, c)
print nonlinear_invert(f, x, y, z)
def generate_data(nobservations, a, b, c, noise_level=0.01):
x, y, z = np.random.random((3, nobservations))
noise = noise_level * np.random.normal(0, noise_level, nobservations)
f = func(x, y, z, a, b, c) + noise
return f, x, y, z
def func(x, y, z, a, b, c):
f = (a**2
+ x * b**2
+ y * a * b * np.cos(c)
+ z * a * b * np.sin(c))
return f
def linear_invert(f, x, y, z):
G = np.vstack([np.ones_like(x), x, y, z]).T
m, _, _, _ = np.linalg.lstsq(G, f)
d, e, f, g = m
a = np.sqrt(d)
b = np.sqrt(e)
c = np.arctan2(g, f) # Note that `c` will be in radians, not degrees
return a, b, c
def nonlinear_invert(f, x, y, z):
# "curve_fit" expects the function to take a slightly different form...
def wrapped_func(observation_points, a, b, c):
x, y, z = observation_points
return func(x, y, z, a, b, c)
xdata = np.vstack([x, y, z])
model, cov = opt.curve_fit(wrapped_func, xdata, f)
return model
main()
You probably want to be using scipy's nonlinear solvers, they're really easy: http://docs.scipy.org/doc/scipy/reference/optimize.nonlin.html