CODEWITHSUNDEEP
pythonmathsympy
Kreuzade
Kreuzade

Reputation: 777

Ignore imaginary roots in sympy

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?

Upvotes: 14

Views: 18710

Answers (6)

kassbohm
kassbohm

Reputation: 41

To get rid of complex numbers with just SymPy (without mpmath) use solve with expand_complex or realroots as mentioned above by @smichr.

from sympy import *
x = var("x")

# With symbol p:
p = var("p", positive=True)
# Without symbol p:
p = 1

a, b, c, d = 1, 15, 10, 1
term = p * (a*x**3 + b*x**2 + c*x + d)

pprint("\nSolution 1:")
sol = solve(term, x)
for x in sol:
    pprint(x)
    tmp = expand_complex(x)
    pprint(tmp)
    pprint(N(tmp))

pprint("\nSolution 2:")
sol = real_roots(term, x)
for x in sol:
    pprint(N(x))

With symbols, use solve with expand_complex.

Upvotes: 1

smichr
smichr

Reputation: 18929

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.

Upvotes: 3

PyRick
PyRick

Reputation: 141

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

Upvotes: 10

asmeurer
asmeurer

Reputation: 91450

If you set x to be real, SymPy will only give you the real solutions

x = Symbol('x', real=True)
solve(..., x)

Upvotes: 16

Kreuzade
Kreuzade

Reputation: 777

As Krastonov had mentioned mpmath provided an easier method:

y = polyroots([int(row["scaleA"]), int(row["scaleB"]), int(row["scaleC"]), int(row["scaleD"])-value])
for root in y:
   if "j" not in str(root):
       value = root

Upvotes: 2

Kreuzade
Kreuzade

Reputation: 777

I managed to simply ignore solutions containing the character "I" and used .evalf() to evaluate the expression. The code is now:

    x = Symbol('x')
    y = solve(int(row["scaleA"])*x**3 + int(row["scaleB"])*x**2 + int(row["scaleC"])*x + int(row["scaleD"]), x)
    for root in y:
        if "I" not in str(root):
            print("This One:" + str(root.evalf()))

Upvotes: -3

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