Given a collection of multivariate inequalities (field of real numbers) I
would like to know if the set defined is connected. Does Sage have a easy
way through qepcad to answer such a question?
Documentation shows that there is a function connected_subset
I tried installing qepcad with
sudo sage -i qepcad
Linking failed with an error about the following file missing
/usr/lib/sagemath/spkg/build/qepcad-1.50/src/saclib2.2.0/lib/saclibo.a
The tail end of the error output log:
Linking the optimized program..
g++ -O4
5:25:54 AM UTC-5, Cary Cherng wrote:
I tried installing qepcad with
sudo sage -i qepcad
I assume you are on some brand of Linux - more processor, etc., info would
be helpful. On my Mac it fails with a completely different error, in
trying to build saclib. (Because `uname` is given
mkproto mkmake and mklib are tsch scripts.
sudo apt-get install tsch solved the problem.
On Thursday, March 7, 2013 5:06:02 PM UTC-8, Cary Cherng wrote:
Machine information:
Linux ccherng 3.5.0-17-generic #28-Ubuntu SMP Tue Oct 9 19:32:08 UTC 2012
i686 athlon i686 GNU/Linux
Upon taking
, p2 0],[g1,g2])
On Monday, December 3, 2012 12:49:19 AM UTC-8, P Purkayastha wrote:
On 12/03/2012 09:44 AM, Cary Cherng wrote:
I tried using solve_ineq in the notebook in the simple way below and got
an error. It seems to be related to
http://trac.sagemath.org/sage_trac/ticket/11520
In the below why does solve_ineq called with the inequalities t1 = t2 , t1
t2 not return [ ], but the other invocations of solve_ineq return the
empty set as [ ] ?
sage: g1,g2 = var('g1,g2')
sage: t1 = g1^2*g2^2
sage: t2 = g1^2*g2
sage: solve_ineq([t1 = t2 , t1 t2],[g1,g2])
[[g1 == 0, 1 g2,
= var('g1,g2')
sage: t1 = large polynomial
sage: t2 = large polynomial
sage: solve_ineq([t1 = t2 , t1 t2],[g1,g2])
# output that is large and impossible to tell visually that it is the empty
set
On Monday, December 3, 2012 9:05:57 PM UTC-8, P Purkayastha wrote:
On 12/04/2012 08:51 AM, Cary
I tried using solve_ineq in the notebook in the simple way below and got an
error. It seems to be related to
http://trac.sagemath.org/sage_trac/ticket/11520
Is there a workaround?
R.g1,g2 = PolynomialRing(QQ)
solve_ineq([g1 g2],[g1,g2])
Traceback (most recent call last):
File stdin, line 1,
R.g17,g19,g27,g28,g38,g39,g47,g49,g57,g58,g68,g69 =
PolynomialRing(QQ)
Eventually I compute a polynomial p with something like
p = p1 / q.determinant()
Sage gives p with type fraction field. How do I cast p back to the
polynomial ring so I can call degree() on it?
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That worked.
On Oct 23, 7:41 pm, John H Palmieri jhpalmier...@gmail.com wrote:
On Oct 23, 7:20 pm, Cary Cherng cche...@gmail.com wrote:
R.g17,g19,g27,g28,g38,g39,g47,g49,g57,g58,g68,g69 =
PolynomialRing(QQ)
Eventually I compute a polynomial p with something like
p = p1 / q.determinant
I have a sage script that ultimately creates a python list called MMv
of length 35354. Each element is a list of length 55. This is in
effect a 35354 by 55 matrix. Print statements show that when I run my
script with load two.sage it gets stuck at taking this list and
creating a matrix. I am using
for this
specific case).
regards
john perry
On Sep 8, 1:57 am, Cary Cherng cche...@gmail.com wrote:
I am not familiar with algebraic geometry or its terminology and new
to sage.
p_1,...p_n and q are elements of Z[x_1,...,x_n]. In my context I have
some evidence that q can be written
nevermind I solved my problem.
On Sep 7, 5:49 pm, Cary Cherng cche...@gmail.com wrote:
This works but is too slow for more complicated examples. Is there a
way to speed up x in I for much bigger examples? Or does this
already use the fastest algorithm based on groebner basis or something
else
I am not familiar with algebraic geometry or its terminology and new
to sage.
p_1,...p_n and q are elements of Z[x_1,...,x_n]. In my context I have
some evidence that q can be written as something like q = p_1*p_2
+ ... + p_5*p_6. In other words q is a degree 2 polynomial in the
p_i's. Can Sage
-0700 (PDT), Cary Cherng cche...@gmail.com
wrote:
Given p_i and q in Q[x_1,...,x_n] I want to see if q is in the ideal
(p1,...,pm). Does sage have easy support for this?
Is this what you're looking for:
sage: R.x, y = QQ[]
sage: I = R.ideal(x^2, y)
sage: x^2*y+y^2 in I
True
sage: x in I
Given p_i and q in Q[x_1,...,x_n] I want to see if q is in the ideal
(p1,...,pm). Does sage have easy support for this?
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I should have given the original full context. These polynomials P,Q,
and p are all in Z[x1,...,xn]. They are all multivariate.
On Sep 3, 8:54 pm, Cary Cherng cche...@gmail.com wrote:
I have a rational function P(x)/Q(x) with numerators and denominators
of very large degree. From the context I
division, but should be
in the manual or help.
Robert
On 4 zář, 05:54, Cary Cherng cche...@gmail.com wrote:
I have a rational function P(x)/Q(x) with numerators and denominators
of very large degree. From the context I know that a certain
polynomial p(x) should divide the denominator. If I
for the denominator?
On Sep 4, 2:06 pm, Justin C. Walker jus...@mac.com wrote:
On Sep 4, 2010, at 13:54 , Cary Cherng wrote:
And thats another problem. How do I tell sage to give me the
denominator of this rational function?
In general, the .denominator and .numerator methods will (should) give
Ok i think I've resolved my problems by avoiding var for declaring
variables and instead using
R.g17,g19,g27,g29,g37,g38,g47,g48,g58,g59,g68,g69 =
PolynomialRing(QQ)
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I have a rational function P(x)/Q(x) with numerators and denominators
of very large degree. From the context I know that a certain
polynomial p(x) should divide the denominator. If I multiply the
numerator by p(x) giving p(x)*P(x)/Q(x) how do I get sage to cancel
p(x) with the factor in Q(x)? In
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