Hi Robert, On 14 Mrz., 23:29, Robert Bradshaw <rober...@math.washington.edu> wrote: > > Yes, but the fact that f(2,3) has a different parent than f(x=2,y=3) > > has a high probability of being troublesome. > > Given f in R[x,y], I think f(x=a, y=b) should do exactly the same > thing as f(a,b). The parent should be the same as R.base_ring()(0) + a > + b.
+1. I think that f.__call__(*args,**kwds) of a polynomial f in variables x,y,z should behave like a Python function of signature f(x,y,z,**kwds). In particular, f(1,2,z=3, foo='bar') should do exactly the same as f(1,2,3,foo='bar'). Note that currently the situation is not very consistent. On the one hand, we already have the behaviour that you wish: sage: R.<x,y> = ZZ[] sage: P.<a,b> = QQ[] sage: f = x+y sage: parent(f(a,b)) Multivariate Polynomial Ring in a, b over Rational Field sage: parent(f(x=a,y=b)) Multivariate Polynomial Ring in a, b over Rational Field sage: parent(f(1,b)) Multivariate Polynomial Ring in a, b over Rational Field But on the other hand, we have the inconsistency that started this thread: sage: parent(f(x=1,y=2)) Multivariate Polynomial Ring in x, y over Integer Ring sage: parent(f(1,2)) Integer Ring > The difficult case is what to do for f(x=5). Should that be the same > as f(x=5, y=y) or a univariate polynomial? I tend to believe that if you evaluate k variables of a polynomial with n variables then the result should be a polynomial with n-k variables. That having said, I wonder whether one should really make a special case for k==n: Should the result be an element of the base ring, or should it actually be a polynomial in zero variables? Cheers, Simon -- To post to this group, send an email to sage-devel@googlegroups.com To unsubscribe from this group, send an email to sage-devel+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-devel URL: http://www.sagemath.org
