How to solve underdetermined systems of polynomial equations?Solve a linear system with more variables than...
Cycles on the torus
Leveling the sagging side of the home
I am the person who abides by rules, but breaks the rules. Who am I?
Can I take the the bonus-action attack from Two-Weapon Fighting without taking the Attack action?
I reported the illegal activity of my boss to his boss. My boss found out. Now I am being punished. What should I do?
How should I solve this integral with changing parameters?
Is there stress on two letters on the word стоят
Create chunks from an array
Will expression retain the same definition if particle is changed?
Converting from "matrix" data into "coordinate" data
What does *dead* mean in *What do you mean, dead?*?
Difference between `nmap local-IP-address` and `nmap localhost`
The (Easy) Road to Code
How to educate team mate to take screenshots for bugs with out unwanted stuff
Movie: boy escapes the real world and goes to a fantasy world with big furry trolls
Use Mercury as quenching liquid for swords?
What do you call someone who likes to pick fights?
School performs periodic password audits. Is my password compromised?
"If + would" conditional in present perfect tense
What is the "determinant" of two vectors?
Why restrict private health insurance?
Is there a logarithm base for which the logarithm becomes an identity function?
Translation of 答えを知っている人はいませんでした
Is it a Cyclops number? "Nobody" knows!
How to solve underdetermined systems of polynomial equations?
Solve a linear system with more variables than equationsHow to solve this particular system of linear equations?How to find value of an unknown in matrix to make system of linear equations consistentSolve the given system of equationsSystems of equations with unknown constantSolving systems of linear equations involving moduloConfusion on linear algebra theorem of systems of equationsFinding the basic and free variables of this matrixFinding solution for a linear system(see below)Linear system of equations and distinct solutions
$begingroup$
I am trying to solve under determined simultaneous non - linear equations, where the variables are multiplied, but the power of the variables is always 1, is there a formal way doing it?
For eg: x + y + z = 6, xyz = 6
I was previously using reduced row Echelon form, but now that is less useful, is there a counterpart of Echelon matrix form for the above kind of equations?
systems-of-equations
edited yesterday
Rodrigo de Azevedo
13.1k41960
asked yesterday
VARUN.N RAOVARUN.N RAO
33
$endgroup$
|
show 1 more comment
$begingroup$
I am trying to solve under determined simultaneous non - linear equations, where the variables are multiplied, but the power of the variables is always 1, is there a formal way doing it?
For eg: x + y + z = 6, xyz = 6
I was previously using reduced row Echelon form, but now that is less useful, is there a counterpart of Echelon matrix form for the above kind of equations?
systems-of-equations
edited yesterday
Rodrigo de Azevedo
13.1k41960
asked yesterday
VARUN.N RAOVARUN.N RAO
33
$endgroup$
$begingroup$
@RodrigodeAzevedo No powers greater than 1 are used, the variables are multiplied that's all
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
Then you're in the realm of algebraic geometry. Solution sets will be algebraic curves, surfaces, etc. You can parameterize them.
$endgroup$
– Rodrigo de Azevedo
yesterday
$begingroup$
@RodrigodeAzevedo is there a name for solving them formally? some method name or something on which I can read on
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
I'm basically trying code a program, where if you give the equation it generates the family of possible answers, so I need something like a ready made formula
$endgroup$
– VARUN.N RAO
yesterday
1
$begingroup$
I don't want to discourage you, but this is a really ambitious task in the general setting.
$endgroup$
– Yves Daoust
yesterday
|
show 1 more comment
$begingroup$
I am trying to solve under determined simultaneous non - linear equations, where the variables are multiplied, but the power of the variables is always 1, is there a formal way doing it?
For eg: x + y + z = 6, xyz = 6
I was previously using reduced row Echelon form, but now that is less useful, is there a counterpart of Echelon matrix form for the above kind of equations?
systems-of-equations
edited yesterday
Rodrigo de Azevedo
13.1k41960
asked yesterday
VARUN.N RAOVARUN.N RAO
33
$endgroup$
I am trying to solve under determined simultaneous non - linear equations, where the variables are multiplied, but the power of the variables is always 1, is there a formal way doing it?
For eg: x + y + z = 6, xyz = 6
I was previously using reduced row Echelon form, but now that is less useful, is there a counterpart of Echelon matrix form for the above kind of equations?
systems-of-equations
systems-of-equations
edited yesterday
Rodrigo de Azevedo
13.1k41960
asked yesterday
VARUN.N RAOVARUN.N RAO
33
edited yesterday
Rodrigo de Azevedo
13.1k41960
asked yesterday
VARUN.N RAOVARUN.N RAO
33
edited yesterday
Rodrigo de Azevedo
13.1k41960
edited yesterday
Rodrigo de Azevedo
13.1k41960
edited yesterday
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked yesterday
VARUN.N RAOVARUN.N RAO
33
asked yesterday
VARUN.N RAOVARUN.N RAO
33
asked yesterday
VARUN.N RAOVARUN.N RAO
33
33
$begingroup$
@RodrigodeAzevedo No powers greater than 1 are used, the variables are multiplied that's all
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
Then you're in the realm of algebraic geometry. Solution sets will be algebraic curves, surfaces, etc. You can parameterize them.
$endgroup$
– Rodrigo de Azevedo
yesterday
$begingroup$
@RodrigodeAzevedo is there a name for solving them formally? some method name or something on which I can read on
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
I'm basically trying code a program, where if you give the equation it generates the family of possible answers, so I need something like a ready made formula
$endgroup$
– VARUN.N RAO
yesterday
1
$begingroup$
I don't want to discourage you, but this is a really ambitious task in the general setting.
$endgroup$
– Yves Daoust
yesterday
|
show 1 more comment
$begingroup$
@RodrigodeAzevedo No powers greater than 1 are used, the variables are multiplied that's all
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
Then you're in the realm of algebraic geometry. Solution sets will be algebraic curves, surfaces, etc. You can parameterize them.
$endgroup$
– Rodrigo de Azevedo
yesterday
$begingroup$
@RodrigodeAzevedo is there a name for solving them formally? some method name or something on which I can read on
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
I'm basically trying code a program, where if you give the equation it generates the family of possible answers, so I need something like a ready made formula
$endgroup$
– VARUN.N RAO
yesterday
1
$begingroup$
I don't want to discourage you, but this is a really ambitious task in the general setting.
$endgroup$
– Yves Daoust
yesterday
$begingroup$
@RodrigodeAzevedo No powers greater than 1 are used, the variables are multiplied that's all
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
@RodrigodeAzevedo No powers greater than 1 are used, the variables are multiplied that's all
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
Then you're in the realm of algebraic geometry. Solution sets will be algebraic curves, surfaces, etc. You can parameterize them.
$endgroup$
– Rodrigo de Azevedo
yesterday
$begingroup$
Then you're in the realm of algebraic geometry. Solution sets will be algebraic curves, surfaces, etc. You can parameterize them.
$endgroup$
– Rodrigo de Azevedo
yesterday
$begingroup$
@RodrigodeAzevedo is there a name for solving them formally? some method name or something on which I can read on
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
@RodrigodeAzevedo is there a name for solving them formally? some method name or something on which I can read on
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
I'm basically trying code a program, where if you give the equation it generates the family of possible answers, so I need something like a ready made formula
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
I'm basically trying code a program, where if you give the equation it generates the family of possible answers, so I need something like a ready made formula
$endgroup$
– VARUN.N RAO
yesterday
1
1
$begingroup$
I don't want to discourage you, but this is a really ambitious task in the general setting.
$endgroup$
– Yves Daoust
yesterday
$begingroup$
I don't want to discourage you, but this is a really ambitious task in the general setting.
$endgroup$
– Yves Daoust
yesterday
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
When you have too many unknowns, an option is to consider some of them as free parameters and solve for the remaining ones. In the given example, solve
$$begin{cases}x+y+z=6,\xyz=6end{cases}$$ where $z$ is assumed to be known.
We can eliminate $y$ by multiplying the first equation by $xz$,
$$begin{cases}x^2z+xyz+xz^2=6xz,\xyz=6end{cases}$$
which leaves us
$$x^2z+x(z^2-6z)+6=0,$$ quadratic in $x$.
This simple trick turns an indeterminate system into a determinate one, but the real problem of solving the determinate one remains.
answered yesterday
Yves DaoustYves Daoust
130k676227
$endgroup$
add a comment |
$begingroup$
Hint: With $$z=frac{6}{xy}$$ we get
$$x^2y+xy^2-6xy+6=0$$ or
$$y^2+y(6+x)+frac{6}{x}=0$$ so we get
$$y_{1,2}=-frac{6+x}{2}pmsqrt{left(frac{6+x}{2}right)^2-frac{6}{x}}$$
The solution is a curve.
And a Groebner bases is given by
$$6-6yz+y^2z+yz^2,x+y+z-6$$
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3139872%2fhow-to-solve-underdetermined-systems-of-polynomial-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When you have too many unknowns, an option is to consider some of them as free parameters and solve for the remaining ones. In the given example, solve
$$begin{cases}x+y+z=6,\xyz=6end{cases}$$ where $z$ is assumed to be known.
We can eliminate $y$ by multiplying the first equation by $xz$,
$$begin{cases}x^2z+xyz+xz^2=6xz,\xyz=6end{cases}$$
which leaves us
$$x^2z+x(z^2-6z)+6=0,$$ quadratic in $x$.
This simple trick turns an indeterminate system into a determinate one, but the real problem of solving the determinate one remains.
answered yesterday
Yves DaoustYves Daoust
130k676227
$endgroup$
add a comment |
$begingroup$
When you have too many unknowns, an option is to consider some of them as free parameters and solve for the remaining ones. In the given example, solve
$$begin{cases}x+y+z=6,\xyz=6end{cases}$$ where $z$ is assumed to be known.
We can eliminate $y$ by multiplying the first equation by $xz$,
$$begin{cases}x^2z+xyz+xz^2=6xz,\xyz=6end{cases}$$
which leaves us
$$x^2z+x(z^2-6z)+6=0,$$ quadratic in $x$.
This simple trick turns an indeterminate system into a determinate one, but the real problem of solving the determinate one remains.
answered yesterday
Yves DaoustYves Daoust
130k676227
$endgroup$
add a comment |
$begingroup$
When you have too many unknowns, an option is to consider some of them as free parameters and solve for the remaining ones. In the given example, solve
$$begin{cases}x+y+z=6,\xyz=6end{cases}$$ where $z$ is assumed to be known.
We can eliminate $y$ by multiplying the first equation by $xz$,
$$begin{cases}x^2z+xyz+xz^2=6xz,\xyz=6end{cases}$$
which leaves us
$$x^2z+x(z^2-6z)+6=0,$$ quadratic in $x$.
This simple trick turns an indeterminate system into a determinate one, but the real problem of solving the determinate one remains.
answered yesterday
Yves DaoustYves Daoust
130k676227
$endgroup$
When you have too many unknowns, an option is to consider some of them as free parameters and solve for the remaining ones. In the given example, solve
$$begin{cases}x+y+z=6,\xyz=6end{cases}$$ where $z$ is assumed to be known.
We can eliminate $y$ by multiplying the first equation by $xz$,
$$begin{cases}x^2z+xyz+xz^2=6xz,\xyz=6end{cases}$$
which leaves us
$$x^2z+x(z^2-6z)+6=0,$$ quadratic in $x$.
This simple trick turns an indeterminate system into a determinate one, but the real problem of solving the determinate one remains.
answered yesterday
Yves DaoustYves Daoust
130k676227
answered yesterday
Yves DaoustYves Daoust
130k676227
answered yesterday
Yves DaoustYves Daoust
130k676227
answered yesterday
Yves DaoustYves Daoust
130k676227
130k676227
add a comment |
add a comment |
$begingroup$
Hint: With $$z=frac{6}{xy}$$ we get
$$x^2y+xy^2-6xy+6=0$$ or
$$y^2+y(6+x)+frac{6}{x}=0$$ so we get
$$y_{1,2}=-frac{6+x}{2}pmsqrt{left(frac{6+x}{2}right)^2-frac{6}{x}}$$
The solution is a curve.
And a Groebner bases is given by
$$6-6yz+y^2z+yz^2,x+y+z-6$$
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
$endgroup$
add a comment |
$begingroup$
Hint: With $$z=frac{6}{xy}$$ we get
$$x^2y+xy^2-6xy+6=0$$ or
$$y^2+y(6+x)+frac{6}{x}=0$$ so we get
$$y_{1,2}=-frac{6+x}{2}pmsqrt{left(frac{6+x}{2}right)^2-frac{6}{x}}$$
The solution is a curve.
And a Groebner bases is given by
$$6-6yz+y^2z+yz^2,x+y+z-6$$
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
$endgroup$
add a comment |
$begingroup$
Hint: With $$z=frac{6}{xy}$$ we get
$$x^2y+xy^2-6xy+6=0$$ or
$$y^2+y(6+x)+frac{6}{x}=0$$ so we get
$$y_{1,2}=-frac{6+x}{2}pmsqrt{left(frac{6+x}{2}right)^2-frac{6}{x}}$$
The solution is a curve.
And a Groebner bases is given by
$$6-6yz+y^2z+yz^2,x+y+z-6$$
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
$endgroup$
Hint: With $$z=frac{6}{xy}$$ we get
$$x^2y+xy^2-6xy+6=0$$ or
$$y^2+y(6+x)+frac{6}{x}=0$$ so we get
$$y_{1,2}=-frac{6+x}{2}pmsqrt{left(frac{6+x}{2}right)^2-frac{6}{x}}$$
The solution is a curve.
And a Groebner bases is given by
$$6-6yz+y^2z+yz^2,x+y+z-6$$
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
edited yesterday
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
77.2k42866
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3139872%2fhow-to-solve-underdetermined-systems-of-polynomial-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
@RodrigodeAzevedo No powers greater than 1 are used, the variables are multiplied that's all
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
Then you're in the realm of algebraic geometry. Solution sets will be algebraic curves, surfaces, etc. You can parameterize them.
$endgroup$
– Rodrigo de Azevedo
yesterday
$begingroup$
@RodrigodeAzevedo is there a name for solving them formally? some method name or something on which I can read on
$endgroup$
– VARUN.N RAO
yesterday
$begingroup$
I'm basically trying code a program, where if you give the equation it generates the family of possible answers, so I need something like a ready made formula
$endgroup$
– VARUN.N RAO
yesterday
1
$begingroup$
I don't want to discourage you, but this is a really ambitious task in the general setting.
$endgroup$
– Yves Daoust
yesterday