Le samedi 24 mai 2014 16:29:38 UTC+2, Tom Harris a écrit :
> Now I have some code to generate the polynomial which I am interested in,
> I store it as p:
>
> p = (output of some functions)
>
> ( p is ((x1^3 - 2*x1*x2 + x3)*c1^2 - (x1*x2 - x3)*c1 + x3)*c2^2 + x1^3 +
> c1^2*x3 - (x1*x2 - x3)*c1
Hi Tom,
Your code works perfectly in Sage 6.2 on Mac
R. = PolynomialRing(ZZ,3)
C. = PolynomialRing(R,2)
Sym = SymmetricFunctions(R)
e = Sym.elementary()
def ElemSym(p):
# checks whether a polynomial is symmetric (coefficients in ZZ[l1,l2,l3])
f = Sym.from_polynomial(p)
return e(f)
p =
+1
On 26 nov, 02:10, Dima Pasechnik wrote:
> Just wonder if anything was actually done in this direction
> (need to compute with some symmetric polynomials now...)
> Dima
>
>
>
>
>
>
>
> On Wednesday, 21 October 2009 01:38:51 UTC+8, Mike Hansen wrote:
>
> > On Wed, Oct 21, 2009 at 12:33 AM, Pierr
Just wonder if anything was actually done in this direction
(need to compute with some symmetric polynomials now...)
Dima
On Wednesday, 21 October 2009 01:38:51 UTC+8, Mike Hansen wrote:
>
> On Wed, Oct 21, 2009 at 12:33 AM, Pierre wrote:
> > thanks for this. I thought about sage this afternoon a
The following works (at least in sage-4.4.4):
sage: R. = QQ[]
sage: SF = SymmetricFunctions(QQ)
sage: SF.from_polynomial(x^2 + y^2 + z^2)
m[2]
sage: SF.from_polynomial(x^2 + y^2)
...
ValueError: x^2 + y^2 is not a symmetric polynomial
Cheers,
Jason
On 07/20/2010 09:23 PM, kcrisman wrote:
> It do
It doesn't seem there is one at this point, though there is a lot of
stuff for using them. This is now
http://trac.sagemath.org/sage_trac/ticket/9558
.
- kcrisman
On Jul 20, 7:11 am, Epsilon wrote:
> Hi,
>
> I would like to know if there are any function that says if a
> polynomial is or not s
On Wed, Oct 21, 2009 at 12:33 AM, Pierre wrote:
> thanks for this. I thought about sage this afternoon and decided the
> easiest would be to pay a visit to david green and yourself in Jena
> and pick your brains... when i checked your homepage just to be
> certain and found out you're in Galway n
hi simon,
thanks for this. I thought about sage this afternoon and decided the
easiest would be to pay a visit to david green and yourself in Jena
and pick your brains... when i checked your homepage just to be
certain and found out you're in Galway now ! Damn. (I also hate the
fact that i've mis
Hi Pierre!
On Oct 20, 2:44 pm, Pierre wrote:
> A priori, the two computations (one with I, one with J) are different.
Yes, probably I was confusing it with a different problem, namely to
find just *some element* in the preimage of an element (if it exists),
but not the complete preimage. Here,
> I guess there will be many people who can offer advise. Just start hacking
> away and others (me included) can help you out once you have concrete
> questions (probably best on [sage-devel])
that's been the plan for years now. Haven't started "hacking" yet...
it probably doesn't help that i hav
> (ii) add the functionality to SAGE : a lot of trouble as far as i'm
> concerned.
>
> Hey if anyone wants to take me through the steps for (ii)... you
> wouldn't believe how little i know.
I guess there will be many people who can offer advise. Just start hacking
away and others (me included)
> However, I am not sure if it is efficient to use these commands often.
> Suppose f is a map from basering to some other ring R, and I, J are
> ideals in R. Then, preimage(R,f,I) computes the preimage of I under f.
> I guess internally some Groebner basis computation is done. When you
> then do p
Hi Martin, hi Pierre!
On Oct 20, 11:25 am, Martin Albrecht
wrote:
> > speaking of shorthands, does SAGE have a ready-made function that
> > computes the kernel of a ring map from a quotient of a polynomial ring
> > to another such quotient ? (straightforward groebner basis computation
> > again)
> speaking of shorthands, does SAGE have a ready-made function that
> computes the kernel of a ring map from a quotient of a polynomial ring
> to another such quotient ? (straightforward groebner basis computation
> again)
I don't think so but it should be relatively easy to add (hint, hint :))
sorry for the late reply. Your code works just fine Marshall, thanks !
I was thinking of writing something similar, but i didn't know the
shorthand elimination_ideal, sweet.
speaking of shorthands, does SAGE have a ready-made function that
computes the kernel of a ring map from a quotient of a po
Groebner bases seem to do this relatively quickly. Here's your last
example done in a crude way, seems almost instantaneous.
R. = PolynomialRing(QQ,order = TermOrder
('degrevlex',4)+TermOrder('degrevlex',3))
foo= (x0 + x1 + x2 + x3)^3
sym = []
xvars = [x0,x1,x2,x3]
for i in range(1,4):
temp
On Mon, Oct 19, 2009 at 8:47 AM, Pierre wrote:
>
> Thanks. This works, but it is so very slow :
>
> sage: foo= (x0 + x1 + x2 + x3)^1;
> sage.libs.symmetrica.all.t_POLYNOM_ELMSYM( foo )
> e[1] #immediate
>
> sage: foo= (x0 + x1 + x2 + x3)^2;
> sage.libs.symmetrica.all.t_POLYNOM_ELMSYM( foo )
Thanks. This works, but it is so very slow :
sage: foo= (x0 + x1 + x2 + x3)^1;
sage.libs.symmetrica.all.t_POLYNOM_ELMSYM( foo )
e[1] #immediate
sage: foo= (x0 + x1 + x2 + x3)^2;
sage.libs.symmetrica.all.t_POLYNOM_ELMSYM( foo )
e[1, 1] #also immediate
sage: foo= (x0 + x1 + x2 + x3)^3;
sage.
Hello,
On Mon, Oct 19, 2009 at 9:13 PM, Pierre wrote:
>
> hi,
>
> i've got some polynomial which happens to be symmetrical. Is there a
> quick and easy way to write it in terms of elementary symmetric
> functions ?
Currently, it's not the most straightforward:
sage: e = SFAElementary(QQ)
sage:
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