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2018-08-08 03:27:38 -0600 received badge  Popular Question (source)
2015-01-13 16:01:15 -0600 commented answer Can I get a polynomial type from a symbolic expression type?

Yup. I didn't know about this P() syntax. Thanks a lot!

2015-01-13 16:01:03 -0600 received badge  Scholar (source)
2015-01-13 14:45:46 -0600 asked a question Can I get a polynomial type from a symbolic expression type?

I am doing some things to a symbolic variable C...

sage: C = var('C')
sage: # ... do a whole bunch of stuff with C as a symbolic variable,
sage: # and at some point get
sage: g = solve(f, C)
sage: print(g)
[0 == 25*C^5 - 325*C^4 + 16276*C^3 - 228488*C^2 - 5940688]
sage: # implicit solution is polynomial in C

... resulting in a polynomial in C that is typed as a symbolic expression:

sage: print g[0].rhs()
25*C^5 - 325*C^4 + 16276*C^3 - 228488*C^2 - 5940688
sage: type(g[0].rhs())
<type 'sage.symbolic.expression.Expression'>

I need to call quo_rem on this expression, so I need it to be a

<type 'sage.rings.polynomial.polynomial_rational_flint.Polynomial_rational_flint'>

I can copy and paste the output of print(g[0].rhs()) into something new and get

sage: P.<C> = PolynomialRing(QQ)
sage: p0 = 25*C^5 - 325*C^4 + 16276*C^3 - 228488*C^2 - 5940688
sage: type(p0)
<type 'sage.rings.polynomial.polynomial_rational_flint.Polynomial_rational_flint'>

and now I can call quo_rem on p0.

How can I get the polynomial type out of the symbolic expression without having to resort to copy and paste?