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
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
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
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
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 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 for
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 ;
and I
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
ideal
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
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
It looks right to me.
I am not a native English speaker so I could be (very) wrong, but I
understand that the comparison x2 is evaluated, which is completely true,
independently if the condition is evaluated as True or False. In fact, next
lines tell why x2 is evaluated False and that h(x)
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
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