On Tuesday, June 25, 2019 at 10:03:03 AM UTC+2, Peter Luschny wrote:
>
> How that? Look at the output above. Sage *knows* that the terms of the sum
> are polynomials. So it should return the zero of that ring, which is the
> null polynomial.
>
>
Not in the first case, look at what are you passi
On Monday, June 17, 2019 at 2:12:58 PM UTC+2, Peter Luschny wrote:
As I see it the problem is that the sum runs over (0..n-1).
> Thus for n = 0 it returns by convention the integer 0 for the
> empty sum (is this correct?) which of course has no list.
>
> But shouldn't it return the null polynomi
On Monday, June 17, 2019 at 2:12:58 PM UTC+2, Peter Luschny wrote:
As I see it the problem is that the sum runs over (0..n-1).
> Thus for n = 0 it returns by convention the integer 0 for the
> empty sum (is this correct?) which of course has no list.
>
> But shouldn't it return the null polynomia
On Wednesday, February 27, 2019 at 2:52:36 PM UTC+1, Daniel Krenn wrote:
> > I suppose in non-full-dimensional case you still can use
> > P.inequalities() as above,
> > projecting them on the affine hull of P.
>
> Yes, this is the interesting case. The problem then is going back from
> the p
Sage interprets that matrices M acts on row vectors v on the left, v*M so
in fact the method image corresponds to row_space
>From the help of image:
Return the image of the homomorphism on *rows* defined by this matrix.
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On Friday, June 12, 2015 at 11:17:37 AM UTC+2, Néstor wrote:
> Hello,
>
> I've got a rational expression in sage and I would like to convert it to a
> polynomial with coefficients in some fraction field.
>
> More precisely, I've got something like this:
>
> a , x = var( 'a , x' ) ;
> P = x/a ;
>
On Thursday, June 11, 2015 at 5:26:28 PM UTC+2, Phoenix wrote:
> I have two polynomials $p(x)$ and $q(x)$ and I want to know if there are
> roots of the equation $\frac{p'}{p} = \frac{q'}{q}$ in the domain
> $(a,\infinity)$ - where $a = max \{ roots(p),roots(q) \}$
>
> This is the same as asking
On Tuesday, June 9, 2015 at 5:36:01 PM UTC+2, black...@gmx.de wrote:
>
> Thank you,
>
> and i already tried this. In this case it obiously does work but in case i
> have denominators, can u explain me how to solve it?
> for example: K(s/(s+t),s^2*t^2) then i have to calculate the elimination
>
Have you tried using elimination ideals?
K=QQ['s,t,a0,a1,a2']
K.inject_variables()
I = Ideal( a0-s^2, a1-t^2, a2 - (s^2+t^2))
I.elimination_ideal([s,t])
Ideal (a0 + a1 - a2) of Multivariate Polynomial Ring in s, t, a0, a1, a2
over Rational Field
So a2 = a0 + a1
The elimination ideal tells you wh
Within a specific interactive session, you could do the following, when
creating the rings:
sage: R = PowerSeriesRing(GF(2),'t')
sage: F = R.residue_field()
sage: phi = R.hom([0], F)
sage: F.register_coercion(phi)
This way, you are indicating that the morphism phi should be considered a
coercio
It looks right to me.
I am not a native English speaker so I could be (very) wrong, but I
understand that the comparison x<2 is evaluated, which is completely true,
independently if the condition is evaluated as True or False. In fact, next
lines tell why x<2 is evaluated False and that h(x) re
In general, I prefer to put the parameters a_i as variables and then
interpret the results.
Another approach you may try is to work in the field:
GF(2^d)['a_1,a_2,a_3'].fraction_field()['x_1,x_2,x_3']
but then you may encounter specialiation problems with denominators,
another problem is that
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