Can I solve a system of high-degree polynomial equations with the Gröbner basis method?
0
$begingroup$
I have two equations, two unknown (ISP, ch) and one parameter ($n_C$). My aim is to find the values of the unknown according to the parameter. The original rational equations are here :
https://framabin.org/p/?5a4a682c865ca030#GHL5aWQ6hPZ6m0csuQwQ7GSj15UWJxliKIohqkj21gY=
I converted them in polynomial, you can look at them here in a $ LaTeX$ format:
https://v1.overleaf.com/23360386gqfrrchwvyxx
I used singular through sagemath to get the Gröbner basis. Nonetheless, singular crashes to get the variety of the ideal. But
The code I used in sagemath :
pch_simple = final2_simple.simplify_rational().rhs().denominator() * ch - final2_simple.simplify_rational().rhs().numerator() #polynom for ch
pisp_simple = final1_simple.simplify_rational().rhs().denominator() * ISP - final1_simple.simplify_rational().rhs().numerator() #polynom for ch
Rsimple = PolynomialRing(FractionField(PolynomialRing(QQ, names='n_C')), names='ISP,ch') #Multivariate Polynomial Ring in ISP, ch over Fraction Field of Univariate Polynomial Ring in n_C over Rational Field
plist_simple = [Rsimple(pch_simple),Rsimple(pisp_simple)] #list (system) of my polynomials
I_simple = ideal(plist_simple) #the ideal of the system of polynomials
G_simple = I_simple.groebner_basis() #the Groebner basis
len(G_simple) #I have three equations in my Groebner basis
I_simple.dimension() #0, the ideal is zero-dimensional
I_simple.variety() #stuck
I think the problem come from the high degree of my second equation ('pisp_simple'). In this equation, ISP is degree two but ch is degree 15. In the first equation ('pch_simple'), ISP is degree one and ch is degree three.
Must I concede my polynomial is impossible to solve or does alternative methods exist to solve my nonlinear system of equations composed by two polynomials ?
polynomials groebner-basis
asked Dec 14 '18 at 17:23
EtienneEtienne
13
$endgroup$
|
show 8 more comments
0
$begingroup$
I have two equations, two unknown (ISP, ch) and one parameter ($n_C$). My aim is to find the values of the unknown according to the parameter. The original rational equations are here :
https://framabin.org/p/?5a4a682c865ca030#GHL5aWQ6hPZ6m0csuQwQ7GSj15UWJxliKIohqkj21gY=
I converted them in polynomial, you can look at them here in a $ LaTeX$ format:
https://v1.overleaf.com/23360386gqfrrchwvyxx
I used singular through sagemath to get the Gröbner basis. Nonetheless, singular crashes to get the variety of the ideal. But
The code I used in sagemath :
pch_simple = final2_simple.simplify_rational().rhs().denominator() * ch - final2_simple.simplify_rational().rhs().numerator() #polynom for ch
pisp_simple = final1_simple.simplify_rational().rhs().denominator() * ISP - final1_simple.simplify_rational().rhs().numerator() #polynom for ch
Rsimple = PolynomialRing(FractionField(PolynomialRing(QQ, names='n_C')), names='ISP,ch') #Multivariate Polynomial Ring in ISP, ch over Fraction Field of Univariate Polynomial Ring in n_C over Rational Field
plist_simple = [Rsimple(pch_simple),Rsimple(pisp_simple)] #list (system) of my polynomials
I_simple = ideal(plist_simple) #the ideal of the system of polynomials
G_simple = I_simple.groebner_basis() #the Groebner basis
len(G_simple) #I have three equations in my Groebner basis
I_simple.dimension() #0, the ideal is zero-dimensional
I_simple.variety() #stuck
I think the problem come from the high degree of my second equation ('pisp_simple'). In this equation, ISP is degree two but ch is degree 15. In the first equation ('pch_simple'), ISP is degree one and ch is degree three.
Must I concede my polynomial is impossible to solve or does alternative methods exist to solve my nonlinear system of equations composed by two polynomials ?
polynomials groebner-basis
asked Dec 14 '18 at 17:23
EtienneEtienne
13
$endgroup$
1
$begingroup$
It looks like you're trying to solve polynomials exactly. That is only possible for fourth-order on down in the first place. For higher-order polynomials, in general, you're going to have to use a numerical method.
$endgroup$
– Adrian Keister
Dec 14 '18 at 17:36
$begingroup$
I was aiming to solve the polynomial exactly according to my parameter $n_C$. I wanted to plot my variables according the values of my parameter. I was inspired by the article Kubler, F., & Schmedders, K. (2010). Tackling Multiplicity of Equilibria with Gröbner Bases. Operations Research, 58(4-part-2), 1037–1050. doi.org/10.1287/opre.1100.0819 . But in my case, the only possibility is to solve numerically, isn't it ?
$endgroup$
– Etienne
Dec 14 '18 at 17:41
$begingroup$
@Adrian Keister has pointed to the problem. If you have, say a 15th degree in one variable and a 2nd degree in another variable, it is likely that it is not the 15th degree that will occur but even the 30th degree, or the 60th degree or wathever... The probability is 0% that your system is solvable formally. You have to turn (as said by the same Adrian Keister) to a numerical method.
$endgroup$
– Jean Marie
Dec 14 '18 at 17:50
1
$begingroup$
It is would be much better to use mathematical notation to express your equations. At a glance the ones you link to (at a site which pops up a modal dialog asking for donations) are rational in the variable $ch$ and not polynomial. A $LaTeX$ version of the equations (preferably using single character variables) would clarify this point.
$endgroup$
– hardmath
Dec 14 '18 at 18:02
1
$begingroup$
@JeanMarie: You may be a little overly pessimistic. Even a 30th degree polynomial might have a lower degree factor. But it appears to me the equations are rational rather than polynomial, at least in one of the two variables.
$endgroup$
– hardmath
Dec 14 '18 at 18:04
|
show 8 more comments
0
0
0
$begingroup$
I have two equations, two unknown (ISP, ch) and one parameter ($n_C$). My aim is to find the values of the unknown according to the parameter. The original rational equations are here :
https://framabin.org/p/?5a4a682c865ca030#GHL5aWQ6hPZ6m0csuQwQ7GSj15UWJxliKIohqkj21gY=
I converted them in polynomial, you can look at them here in a $ LaTeX$ format:
https://v1.overleaf.com/23360386gqfrrchwvyxx
I used singular through sagemath to get the Gröbner basis. Nonetheless, singular crashes to get the variety of the ideal. But
The code I used in sagemath :
pch_simple = final2_simple.simplify_rational().rhs().denominator() * ch - final2_simple.simplify_rational().rhs().numerator() #polynom for ch
pisp_simple = final1_simple.simplify_rational().rhs().denominator() * ISP - final1_simple.simplify_rational().rhs().numerator() #polynom for ch
Rsimple = PolynomialRing(FractionField(PolynomialRing(QQ, names='n_C')), names='ISP,ch') #Multivariate Polynomial Ring in ISP, ch over Fraction Field of Univariate Polynomial Ring in n_C over Rational Field
plist_simple = [Rsimple(pch_simple),Rsimple(pisp_simple)] #list (system) of my polynomials
I_simple = ideal(plist_simple) #the ideal of the system of polynomials
G_simple = I_simple.groebner_basis() #the Groebner basis
len(G_simple) #I have three equations in my Groebner basis
I_simple.dimension() #0, the ideal is zero-dimensional
I_simple.variety() #stuck
I think the problem come from the high degree of my second equation ('pisp_simple'). In this equation, ISP is degree two but ch is degree 15. In the first equation ('pch_simple'), ISP is degree one and ch is degree three.
Must I concede my polynomial is impossible to solve or does alternative methods exist to solve my nonlinear system of equations composed by two polynomials ?
polynomials groebner-basis
asked Dec 14 '18 at 17:23
EtienneEtienne
13
$endgroup$
I have two equations, two unknown (ISP, ch) and one parameter ($n_C$). My aim is to find the values of the unknown according to the parameter. The original rational equations are here :
https://framabin.org/p/?5a4a682c865ca030#GHL5aWQ6hPZ6m0csuQwQ7GSj15UWJxliKIohqkj21gY=
I converted them in polynomial, you can look at them here in a $ LaTeX$ format:
https://v1.overleaf.com/23360386gqfrrchwvyxx
I used singular through sagemath to get the Gröbner basis. Nonetheless, singular crashes to get the variety of the ideal. But
The code I used in sagemath :
pch_simple = final2_simple.simplify_rational().rhs().denominator() * ch - final2_simple.simplify_rational().rhs().numerator() #polynom for ch
pisp_simple = final1_simple.simplify_rational().rhs().denominator() * ISP - final1_simple.simplify_rational().rhs().numerator() #polynom for ch
Rsimple = PolynomialRing(FractionField(PolynomialRing(QQ, names='n_C')), names='ISP,ch') #Multivariate Polynomial Ring in ISP, ch over Fraction Field of Univariate Polynomial Ring in n_C over Rational Field
plist_simple = [Rsimple(pch_simple),Rsimple(pisp_simple)] #list (system) of my polynomials
I_simple = ideal(plist_simple) #the ideal of the system of polynomials
G_simple = I_simple.groebner_basis() #the Groebner basis
len(G_simple) #I have three equations in my Groebner basis
I_simple.dimension() #0, the ideal is zero-dimensional
I_simple.variety() #stuck
I think the problem come from the high degree of my second equation ('pisp_simple'). In this equation, ISP is degree two but ch is degree 15. In the first equation ('pch_simple'), ISP is degree one and ch is degree three.
Must I concede my polynomial is impossible to solve or does alternative methods exist to solve my nonlinear system of equations composed by two polynomials ?
polynomials groebner-basis
polynomials groebner-basis
asked Dec 14 '18 at 17:23
EtienneEtienne
13
asked Dec 14 '18 at 17:23
EtienneEtienne
13
edited Dec 15 '18 at 10:38
Etienne
asked Dec 14 '18 at 17:23
EtienneEtienne
13
asked Dec 14 '18 at 17:23
EtienneEtienne
13
asked Dec 14 '18 at 17:23
EtienneEtienne
13
13
1
$begingroup$
It looks like you're trying to solve polynomials exactly. That is only possible for fourth-order on down in the first place. For higher-order polynomials, in general, you're going to have to use a numerical method.
$endgroup$
– Adrian Keister
Dec 14 '18 at 17:36
$begingroup$
I was aiming to solve the polynomial exactly according to my parameter $n_C$. I wanted to plot my variables according the values of my parameter. I was inspired by the article Kubler, F., & Schmedders, K. (2010). Tackling Multiplicity of Equilibria with Gröbner Bases. Operations Research, 58(4-part-2), 1037–1050. doi.org/10.1287/opre.1100.0819 . But in my case, the only possibility is to solve numerically, isn't it ?
$endgroup$
– Etienne
Dec 14 '18 at 17:41
$begingroup$
@Adrian Keister has pointed to the problem. If you have, say a 15th degree in one variable and a 2nd degree in another variable, it is likely that it is not the 15th degree that will occur but even the 30th degree, or the 60th degree or wathever... The probability is 0% that your system is solvable formally. You have to turn (as said by the same Adrian Keister) to a numerical method.
$endgroup$
– Jean Marie
Dec 14 '18 at 17:50
1
$begingroup$
It is would be much better to use mathematical notation to express your equations. At a glance the ones you link to (at a site which pops up a modal dialog asking for donations) are rational in the variable $ch$ and not polynomial. A $LaTeX$ version of the equations (preferably using single character variables) would clarify this point.
$endgroup$
– hardmath
Dec 14 '18 at 18:02
1
$begingroup$
@JeanMarie: You may be a little overly pessimistic. Even a 30th degree polynomial might have a lower degree factor. But it appears to me the equations are rational rather than polynomial, at least in one of the two variables.
$endgroup$
– hardmath
Dec 14 '18 at 18:04
|
show 8 more comments
1
$begingroup$
It looks like you're trying to solve polynomials exactly. That is only possible for fourth-order on down in the first place. For higher-order polynomials, in general, you're going to have to use a numerical method.
$endgroup$
– Adrian Keister
Dec 14 '18 at 17:36
$begingroup$
I was aiming to solve the polynomial exactly according to my parameter $n_C$. I wanted to plot my variables according the values of my parameter. I was inspired by the article Kubler, F., & Schmedders, K. (2010). Tackling Multiplicity of Equilibria with Gröbner Bases. Operations Research, 58(4-part-2), 1037–1050. doi.org/10.1287/opre.1100.0819 . But in my case, the only possibility is to solve numerically, isn't it ?
$endgroup$
– Etienne
Dec 14 '18 at 17:41
$begingroup$
@Adrian Keister has pointed to the problem. If you have, say a 15th degree in one variable and a 2nd degree in another variable, it is likely that it is not the 15th degree that will occur but even the 30th degree, or the 60th degree or wathever... The probability is 0% that your system is solvable formally. You have to turn (as said by the same Adrian Keister) to a numerical method.
$endgroup$
– Jean Marie
Dec 14 '18 at 17:50
1
$begingroup$
It is would be much better to use mathematical notation to express your equations. At a glance the ones you link to (at a site which pops up a modal dialog asking for donations) are rational in the variable $ch$ and not polynomial. A $LaTeX$ version of the equations (preferably using single character variables) would clarify this point.
$endgroup$
– hardmath
Dec 14 '18 at 18:02
1
$begingroup$
@JeanMarie: You may be a little overly pessimistic. Even a 30th degree polynomial might have a lower degree factor. But it appears to me the equations are rational rather than polynomial, at least in one of the two variables.
$endgroup$
– hardmath
Dec 14 '18 at 18:04
1
1
$begingroup$
It looks like you're trying to solve polynomials exactly. That is only possible for fourth-order on down in the first place. For higher-order polynomials, in general, you're going to have to use a numerical method.
$endgroup$
– Adrian Keister
Dec 14 '18 at 17:36
$begingroup$
It looks like you're trying to solve polynomials exactly. That is only possible for fourth-order on down in the first place. For higher-order polynomials, in general, you're going to have to use a numerical method.
$endgroup$
– Adrian Keister
Dec 14 '18 at 17:36
$begingroup$
I was aiming to solve the polynomial exactly according to my parameter $n_C$. I wanted to plot my variables according the values of my parameter. I was inspired by the article Kubler, F., & Schmedders, K. (2010). Tackling Multiplicity of Equilibria with Gröbner Bases. Operations Research, 58(4-part-2), 1037–1050. doi.org/10.1287/opre.1100.0819 . But in my case, the only possibility is to solve numerically, isn't it ?
$endgroup$
– Etienne
Dec 14 '18 at 17:41
$begingroup$
I was aiming to solve the polynomial exactly according to my parameter $n_C$. I wanted to plot my variables according the values of my parameter. I was inspired by the article Kubler, F., & Schmedders, K. (2010). Tackling Multiplicity of Equilibria with Gröbner Bases. Operations Research, 58(4-part-2), 1037–1050. doi.org/10.1287/opre.1100.0819 . But in my case, the only possibility is to solve numerically, isn't it ?
$endgroup$
– Etienne
Dec 14 '18 at 17:41
$begingroup$
@Adrian Keister has pointed to the problem. If you have, say a 15th degree in one variable and a 2nd degree in another variable, it is likely that it is not the 15th degree that will occur but even the 30th degree, or the 60th degree or wathever... The probability is 0% that your system is solvable formally. You have to turn (as said by the same Adrian Keister) to a numerical method.
$endgroup$
– Jean Marie
Dec 14 '18 at 17:50
$begingroup$
@Adrian Keister has pointed to the problem. If you have, say a 15th degree in one variable and a 2nd degree in another variable, it is likely that it is not the 15th degree that will occur but even the 30th degree, or the 60th degree or wathever... The probability is 0% that your system is solvable formally. You have to turn (as said by the same Adrian Keister) to a numerical method.
$endgroup$
– Jean Marie
Dec 14 '18 at 17:50
1
1
$begingroup$
It is would be much better to use mathematical notation to express your equations. At a glance the ones you link to (at a site which pops up a modal dialog asking for donations) are rational in the variable $ch$ and not polynomial. A $LaTeX$ version of the equations (preferably using single character variables) would clarify this point.
$endgroup$
– hardmath
Dec 14 '18 at 18:02
$begingroup$
It is would be much better to use mathematical notation to express your equations. At a glance the ones you link to (at a site which pops up a modal dialog asking for donations) are rational in the variable $ch$ and not polynomial. A $LaTeX$ version of the equations (preferably using single character variables) would clarify this point.
$endgroup$
– hardmath
Dec 14 '18 at 18:02
1
1
$begingroup$
@JeanMarie: You may be a little overly pessimistic. Even a 30th degree polynomial might have a lower degree factor. But it appears to me the equations are rational rather than polynomial, at least in one of the two variables.
$endgroup$
– hardmath
Dec 14 '18 at 18:04
$begingroup$
@JeanMarie: You may be a little overly pessimistic. Even a 30th degree polynomial might have a lower degree factor. But it appears to me the equations are rational rather than polynomial, at least in one of the two variables.
$endgroup$
– hardmath
Dec 14 '18 at 18:04
|
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1
$begingroup$
It looks like you're trying to solve polynomials exactly. That is only possible for fourth-order on down in the first place. For higher-order polynomials, in general, you're going to have to use a numerical method.
$endgroup$
– Adrian Keister
Dec 14 '18 at 17:36
$begingroup$
I was aiming to solve the polynomial exactly according to my parameter $n_C$. I wanted to plot my variables according the values of my parameter. I was inspired by the article Kubler, F., & Schmedders, K. (2010). Tackling Multiplicity of Equilibria with Gröbner Bases. Operations Research, 58(4-part-2), 1037–1050. doi.org/10.1287/opre.1100.0819 . But in my case, the only possibility is to solve numerically, isn't it ?
$endgroup$
– Etienne
Dec 14 '18 at 17:41
$begingroup$
@Adrian Keister has pointed to the problem. If you have, say a 15th degree in one variable and a 2nd degree in another variable, it is likely that it is not the 15th degree that will occur but even the 30th degree, or the 60th degree or wathever... The probability is 0% that your system is solvable formally. You have to turn (as said by the same Adrian Keister) to a numerical method.
$endgroup$
– Jean Marie
Dec 14 '18 at 17:50
1
$begingroup$
It is would be much better to use mathematical notation to express your equations. At a glance the ones you link to (at a site which pops up a modal dialog asking for donations) are rational in the variable $ch$ and not polynomial. A $LaTeX$ version of the equations (preferably using single character variables) would clarify this point.
$endgroup$
– hardmath
Dec 14 '18 at 18:02
1
$begingroup$
@JeanMarie: You may be a little overly pessimistic. Even a 30th degree polynomial might have a lower degree factor. But it appears to me the equations are rational rather than polynomial, at least in one of the two variables.
$endgroup$
– hardmath
Dec 14 '18 at 18:04