Identification polynomial

asked 2010-11-14 04:04:49 +0100

czsan
237 ●8 ●14 ●25

I have a function with one parameter. How can I check whether this parameter is a polynomial? (Univariate or multvariate, does'nt count which is the base ring) I want use this information in a multilevel if construction

answered 2010-11-14 06:44:12 +0100

DSM

5242 ●21 ●68 ●112

Will something like this work?

sage: is_polynomial = lambda s: all(s.is_polynomial(v) for v in s.variables())
sage: var("x y z")
(x, y, z)
sage: is_polynomial(x)
True
sage: is_polynomial(3+x+x**7)
True
sage: is_polynomial(3+x*exp(x))
False
sage: is_polynomial(3*x*y)
True
sage: is_polynomial(3*x*y+sin(y))
False

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Thank you, but it is a partial solution, becaus I do'nt know, the parameter has varibales() property or not. For example : a=2 is_polynomial(a) generates an AttributeError. I believed, the .parent() will help me, but there are rings without this property.

czsan ( 2010-11-15 11:07:53 +0100 )edit

Ah, so you don't even know that it's an Expression, then. Maybe casting it would suffice: is_polynomial = lambda s: all(SR(s).is_polynomial(v) for v in SR(s).variables()) but I'm not clear on your use cases (and how much information you want to preserve).

DSM ( 2010-11-15 12:29:32 +0100 )edit

Yes, I don't know, the parameter is a number or a polynom. After some preparation the algoritm is same (for example, gcd, cra). but the preparation and some little aid-functions depend on the parent ring.

czsan ( 2010-11-15 13:28:06 +0100 )edit

So does casting your parameter explicitly into an Expression by wrapping it in SR suffice for your case, or do we need to work harder?

DSM ( 2010-11-15 18:52:19 +0100 )edit

I think it's hard. What do you mean on 'SR suffice'?

czsan ( 2010-11-18 15:08:04 +0100 )edit

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