I want to solve f(x) from
(Eq(2*f(x)+f(1/x),1/x))
the expected output is
f(x) = (2-x^2)/(3*x)
I try
solve((Eq(2*f(x)+f(1/x),1/x)),f(x))
This answer contains f(1/x) : (-x*f(1/x) + 1)/(2*x)
How to get f(x) = (2-x^2)/(3*x) in sympy?
I don't think sympy will solve functional equations like this in the way you want, but you could separate your particular equation into two: 2y+z = 1/x and 2z+y = x where y(1/x) = z(x) and let sympy solve for both y and z:
In [5]: x, y, z = Symbol('x'), Symbol('y'), Symbol('z')
In [6]: solve((Eq(2*y+z,1/x), Eq(2*z+y,x)),y,z)
Out[6]: {y: (-x**2 + 2)/(3*x), z: (2*x**2 - 1)/(3*x)}
So y is the f(x) that you want here.
Related
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.
I have a function [ -4*x/sqrt(1 - (1 - 2*x^2)^2) + 2/sqrt(1 - x^2) ] that I need to evaluate at x=0. However, whenever you graph this function, for some interval of y there are many y-values at x=0. This leads me to think that the (subs) command can only return one y-value. Any help or elaboration on this? Thank you!
Here's my code if it might help:
x = symbols('x')
f = 2*asin(x) # f(x) function
g = acos(1-2*x**2) # g(x) function
eq = diff(f-g) # evaluating the derivative of f(x) - g(x)
eq.subs(x, 0) # substituting 0 for x in the derivative of f(x) - g(x)
After I run the code, it returns NaN, which I assume is because substituting in 0 for x returns not a single number, but a range of numbers.
Here is the graph of the function to be evaluated at x=0:
You should always give SymPy as many assumptions as possible. For example, it can't pull an x**2 out of a sqrt because it thinks x is complex.
A factorization an then a simplification solves the problem. SymPy can't do L'Hopital on eq = A + B since it does not know that both A and B converge. So you have to guide it a little by bringing the fractions together and then simplifying:
from sympy import *
x = symbols('x', real=True)
f = 2*asin(x) # f(x) function
g = acos(1-2*x**2) # g(x) function
eq = diff(f-g) # evaluating the derivative of f(x) - g(x)
eq = simplify(factor(eq))
print(eq)
print(limit(eq, x, 0, "+"))
print(limit(eq, x, 0, "-"))
Outputs:
(-2*x + 2*Abs(x))/(sqrt(1 - x**2)*Abs(x))
0
4
simplify, factor and expand do wonders.
How can I get this to give me x = z*y/a ?
from sympy import *
x,y,a,z = symbols('x y a z')
z = a*x/y
solve(z,x) # returns 0!
# would like to get z*y/a
solve(z,x) correctly returns 0 because your code is effectively asking:
What's the value of x that would cause z to become 0?
What you really want to do (as described here) is solve a*x/y==z which can be done as follows:
from sympy import *
x,y,a,z = symbols('x y a z')
equation = a*x/y
new_eq = solve(equation - z, x) # its value is [y*z/a]
Don't assign z = a*x/y, and don't pass z to solve.
solve(expr, symbol) determines what values of symbol will make expr equal 0. If you want to figure out what value of x makes z equal a*x/y, you want z - a*x/y to equal 0:
solve(z - a*x/y, x)
You do not want to assign z = a*x/y. = means something entirely different from equality.
I think the answer to this question can be of help. Applied to your example, this gives:
>>> from sympy import *
>>> x,y,a,z = symbols('x y a z')
>>> l = z
>>> r = a*x/y
>>> solve(l-r,x)
[y*z/a]
As all the other answers points out the solution,
I would like to emphasize on the use of Eq instances here.
An Eq object represents An equal relation between two objects.
For using an Eq object, your code should look something like this:
In []: a, x, y, z = symbols('a, x, y, z')
In []: foo = Eq(z, a*x/y)
In []: solve(foo, x)
Out[]: [y*z/a]
I've been doing derivatives in sympy, and I didn't know how that would syntactically be written. I tried looking it up, but none of the solutions made sense. For example, if I'm trying to differentiate x**5 + y**2 + z**4 = 8xyz by computation, how would I do that? Would z be a symbol, or a function like in regular derivatives? Thank you.
For two variables you can use idiff.
In your case, the simplest way is to set x and y to be functions of z, like
x = Function('x')(z)
y = Function('y')(z)
Then normal diff(expr, z) will take the derivative correctly.
You commented you used idiff, here is the corresponding code for those who want to know:
from sympy import symbols, idiff, simplify
x, y, z = symbols('x y z')
ex = x**5 + y**2 + z**4 - 8*x*y*z
ex_d = simplify(idiff(ex,(x,y),z))
display(ex_d)
In idiff(ex,(x,y),z), (x,y) are the dependent variables, and z the variable the derivative is being taken with respect to.
I am trying to figure out why sympy.diff does not differentiate sympy polynomials as expected. Normally, sympy.diff works just fine if a symbolic variable is defined and the polynomial is NOT defined using sympy.Poly. However, if the function is defined using sympy.Poly, sympy.diff does not seem to actually compute the derivative. Below is a code sample that shows what I mean:
import sympy as sy
# define symbolic variables
x = sy.Symbol('x')
y = sy.Symbol('y')
# define function WITHOUT using sy.Poly
f1 = x + 1
# define function WITH using sy.Poly
f2 = sy.Poly(x + 1, x, domain='QQ')
# compute derivatives and return results
df1 = sy.diff(f1,x)
df2 = sy.diff(f2,x)
print('f1: ',f1)
print('f2: ',f2)
print('df1: ',df1)
print('df2: ',df2)
This prints the following results:
f1: x + 1
f2: Poly(x + 1, x, domain='QQ')
df1: 1
df2: Derivative(Poly(x + 1, x, domain='QQ'), x)
Why does sympy.diff not know how to differentiate the sympy.Poly version of the polynomial? Is there a way to differentiate the sympy polynomial, or a way to convert the sympy polynomial to the form that allows it to be differentiated?
Note: I tried with different domains (i.e., domain='RR' instead of domain='QQ'), and the output does not change.
This appears to be a bug. You can get around it by calling diff directly on the Poly instance. Ideally calling the function diff from the top level sympy module should yield the same result as calling the method diff.
In [1]: from sympy import *
In [2]: from sympy.abc import x
In [3]: p = Poly(x+1, x, domain='QQ')
In [4]: p.diff(x)
Out[4]: Poly(1, x, domain='QQ')
In [5]: diff(p, x)
Out[5]: Derivative(Poly(x + 1, x, domain='QQ'), x)
In [6]: diff(p, x).doit()
Out[6]: Derivative(Poly(x + 1, x, domain='ZZ'), x)