I'm using sympy to solve a polynomial:
x = Symbol('x')
y = solve(int(row["scaleA"])*x**3 + int(row["scaleB"])*x**2 + int(row["scaleC"])*x + int(row["scaleD"]), x)
y is a list of possible solutions. However, I need to ignore the imaginary ones and only use the real solutions. Also, I would like the solution as a value not an expression. Right now it looks like:
[-2/3 - 55**(1/3)*(-1/2 - sqrt(3)*I/2)/3, -2/3 - 55**(1/3)*(-1/2 + sqrt(3)*I/2)/3, -55**(1/3)/3 - 2/3]
I need the last expression's value (-2.22756). Are there functions in sympy to simplify this?
If you set x
to be real, SymPy will only give you the real solutions
x = Symbol('x', real=True)
solve(..., x)
solve()
doesn’t have a consistent output for various types of solutions, please use solveset(Eq,x,domain=S.Reals)
:
from sympy import ImageSet, S
x = Symbol('x')
y = solveset(int(row["scaleA"])*x**3 + int(row["scaleB"])*x**2+int(row["scaleC"])*x + int(row["scaleD"]), x, domain=S.Reals)
http://docs.sympy.org/latest/modules/solvers/solveset.html
This is exactly the sort of thing that real_roots
is made for and is especially applicable to your case where the coefficients are integers:
x = Symbol('x')
eq = int(row["scaleA"])*x**3 + int(row["scaleB"])*x**2 + int(row["scaleC"])*x + int(row["scaleD"])
y = real_roots(eq, x) # gives [CRootOf(...), ...]
The value of CRootOf instances can be evaluated to whatever precision you need and should not contain any imaginary part. For example,
>>> [i.n(12) for i in real_roots(3*x**3 - 2*x**2 + 7*x - 9, x)]
[1.07951904858]
Note: As I recall, solve will send back roots that it wasn't able to confirm met the assumptions (i.e. if they weren't found to be false for the assumption then they are returned). Also, if you want more consistent output from solve, @PyRick, set the flag dict=True
.
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