Re: [sage-support] backward substitution during solving polynomial equation
Daniel Krenn wrote:
I want to solve polynomial equations and in order
to do so, I do something like:
sage: R.x,y = PolynomialRing(QQ, order='lex')
sage: I = R.ideal([x*y-1, x^2-y^2])
sage: I.groebner_basis()
[x - y^3, y^4 - 1]
and then wrote:
Meanwhile, I found, which seems to do what I want:
sage: I.variety()
[{y: -1, x: -1}, {y: 1, x: 1}]
sage: I.variety(ring=QQbar)
[{y: -1, x: -1}, {y: -1*I, x: 1*I}, {y: 1*I, x: -1*I}, {y: 1, x: 1}]
sage: I.variety(ring=ZZ)
[{y: -1, x: -1}, {y: 1, x: 1}]
On a related note, see the following at ask-sage:
http://ask.sagemath.org/question/8224/system-of-nonlinear-equations/
http://ask.sagemath.org/question/11070/find-algebraic-solutions-to-system-of-polynomial-equations/
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Re: [sage-support] backward substitution during solving polynomial equation
2014-09-01 14:13 UTC+01:00, slelievre samuel.lelie...@gmail.com:
Daniel Krenn wrote:
I want to solve polynomial equations and in order
to do so, I do something like:
sage: R.x,y = PolynomialRing(QQ, order='lex')
sage: I = R.ideal([x*y-1, x^2-y^2])
sage: I.groebner_basis()
[x - y^3, y^4 - 1]
and then wrote:
Meanwhile, I found, which seems to do what I want:
sage: I.variety()
[{y: -1, x: -1}, {y: 1, x: 1}]
sage: I.variety(ring=QQbar)
[{y: -1, x: -1}, {y: -1*I, x: 1*I}, {y: 1*I, x: -1*I}, {y: 1, x: 1}]
sage: I.variety(ring=ZZ)
[{y: -1, x: -1}, {y: 1, x: 1}]
On a related note, see the following at ask-sage:
http://ask.sagemath.org/question/8224/system-of-nonlinear-equations/
http://ask.sagemath.org/question/11070/find-algebraic-solutions-to-system-of-polynomial-equations/
J'ai la flemme de le faire, mais c'est cool que tu fasses les liens
sage-support - ask.sagemath !!
Pour les surfaces a petits carreaux, j'ai surtout du code et la base
de donnees. Cela dit, je suis pas emballe par utiliser le cloud...
C'etait bien les Etats-Unis?
A bientot
Vincent
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Re: [sage-support] backward substitution during solving polynomial equation
2014-09-01 18:56 UTC+01:00, Vincent Delecroix 20100.delecr...@gmail.com:
2014-09-01 14:13 UTC+01:00, slelievre samuel.lelie...@gmail.com:
Daniel Krenn wrote:
I want to solve polynomial equations and in order
to do so, I do something like:
sage: R.x,y = PolynomialRing(QQ, order='lex')
sage: I = R.ideal([x*y-1, x^2-y^2])
sage: I.groebner_basis()
[x - y^3, y^4 - 1]
and then wrote:
Meanwhile, I found, which seems to do what I want:
sage: I.variety()
[{y: -1, x: -1}, {y: 1, x: 1}]
sage: I.variety(ring=QQbar)
[{y: -1, x: -1}, {y: -1*I, x: 1*I}, {y: 1*I, x: -1*I}, {y: 1, x: 1}]
sage: I.variety(ring=ZZ)
[{y: -1, x: -1}, {y: 1, x: 1}]
On a related note, see the following at ask-sage:
http://ask.sagemath.org/question/8224/system-of-nonlinear-equations/
http://ask.sagemath.org/question/11070/find-algebraic-solutions-to-system-of-polynomial-equations/
J'ai la flemme de le faire, mais c'est cool que tu fasses les liens
sage-support - ask.sagemath !!
Pour les surfaces a petits carreaux, j'ai surtout du code et la base
de donnees. Cela dit, je suis pas emballe par utiliser le cloud...
C'etait bien les Etats-Unis?
Sorry!! It was entended to be for Samuel only! Hopefully, nothing very
confidential ;-)
Vincent
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Re: [sage-support] backward substitution during solving polynomial equation
Am 2014-08-29 um 21:25 schrieb Daniel Krenn:
I want to solve polynomial equations and in order to do so, I do
something like:
sage: R.x,y = PolynomialRing(QQ, order='lex')
sage: I = R.ideal([x*y-1, x^2-y^2])
sage: I.groebner_basis()
[x - y^3, y^4 - 1]
Meanwhile, I found, which seems to do what I want:
sage: I.variety()
[{y: -1, x: -1}, {y: 1, x: 1}]
sage: I.variety(ring=QQbar)
[{y: -1, x: -1}, {y: -1*I, x: 1*I}, {y: 1*I, x: -1*I}, {y: 1, x: 1}]
sage: I.variety(ring=ZZ)
[{y: -1, x: -1}, {y: 1, x: 1}]
Daniel
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Re: [sage-support] backward substitution during solving polynomial equation
2014-08-31 11:51 UTC+02:00, Daniel Krenn kr...@aon.at:
Am 2014-08-29 um 21:25 schrieb Daniel Krenn:
I want to solve polynomial equations and in order to do so, I do
something like:
sage: R.x,y = PolynomialRing(QQ, order='lex')
sage: I = R.ideal([x*y-1, x^2-y^2])
sage: I.groebner_basis()
[x - y^3, y^4 - 1]
Meanwhile, I found, which seems to do what I want:
sage: I.variety()
[{y: -1, x: -1}, {y: 1, x: 1}]
sage: I.variety(ring=QQbar)
[{y: -1, x: -1}, {y: -1*I, x: 1*I}, {y: 1*I, x: -1*I}, {y: 1, x: 1}]
sage: I.variety(ring=ZZ)
[{y: -1, x: -1}, {y: 1, x: 1}]
Ho! Much better. Thanks for posting it. I learned something ;-)
Vincent
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Re: [sage-support] backward substitution during solving polynomial equation
(solve seems to be very much an overkill and it is not that transparent in what it does...) Definitely! And I won't even believe the output... I want to solve polynomial equations and in order to do so, I do something like: sage: R.x,y = PolynomialRing(QQ, order='lex') sage: I = R.ideal([x*y-1, x^2-y^2]) sage: I.groebner_basis() [x - y^3, y^4 - 1] What is the Sage command to do this operation, i.e., backwards substituting to find a solution? One possibility (I do not think there is a ready made function) sage: sage: R.x,y = PolynomialRing(QQ, order='lex') sage: sage: I = R.ideal([x*y-1, x^2-y^2]) sage: sage: I.groebner_basis() [x - y^3, y^4 - 1] sage: f1, f2 = I.groebner_basis() sage: Ry.y = PolynomialRing(QQ) sage: Rx.x = PolynomialRing(QQbar) sage: roots_y = Ry(f2).roots(QQbar) sage: print roots_y [(-1, 1), (1, 1), (-1*I, 1), (1*I, 1)] sage: for r,_ in roots_y: :for s,_ in Rx(f1.subs(y=r)).roots(QQbar): :print (s,r) (-1, -1) (1, 1) (1*I, -1*I) (-1*I, 1*I) Vincent -- You received this message because you are subscribed to the Google Groups sage-support group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-support+unsubscr...@googlegroups.com. To post to this group, send email to sage-support@googlegroups.com. Visit this group at http://groups.google.com/group/sage-support. For more options, visit https://groups.google.com/d/optout.
