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System of equations (let me see You!)

  • Thread starter tehno
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Consider the following system of algebraic equations:

[tex]2x+2y+2z-3=0[/tex]
[tex]x^3+3yz^2-u=0[/tex]
[tex]y^3+3x^2z-u=0[/tex]
[tex]z^3+3xy^2-u=0[/tex]

Problem questions:
a)Is the quadruple (1/2,1/2,1/2,1/2) the only solution to the system in [tex]\mathbb{R}_{+}[/tex] ?

b)If the answer to a) is "yes" how to proove that fact;if the answer is "no" how to determine how many other (nonelementar) solutions the system of equations like that will have ?
 

Hurkyl

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This looks like homework. Have you tried anything?
 
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This looks like homework. Have you tried anything?
If you find this problem a homework and trivial than I'm really scared of level you work at.:biggrin:
No friend ,this is not a homework .
That's the challenge to you and everybody else.
(I know the answer just becouse I created that problem but whoever I asked failed to "solve" it so far.:smile:)

EDIT:I kindly suggest problem to be returned where It was posted first,or shifted to General math subforum.
 
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dextercioby

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I'll take a bite.

a) yes
b) cause Maple says so.

Daniel.
 
360
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I'll take a bite.

a) yes
b) cause Maple says so.

Daniel.

Daniel your guess a) is correct but...
in relation to your remark b) may I ask you do you use computer to solve
say quadratic equations too?You know computers can do these tasks extremly well :biggrin:

I say there is a strict mathematical way to demonstrate the only positive solution to the system is the one I gave above.

Next person please.
 
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Gib Z

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Jesus techno you sick son of a *****, I expanded (x+y+z)^3 and all I've got it reduced to is [tex]u=\frac{9}{8} -(x^2y+2xyz+y^2z+xz^2)[/tex] and it took a bloody long time...anyone got a link to an expander?
 
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I don't want to spoil the party and I'll just give you two hints for this "homework" that might help:

1)The system is impossible to solve explicitely,by any kind of algebraic manipulation,substitution or transformation.
2)What is given is the only positive solution to it,and what should one notice is that there is certain level of symmetry of 3 variable function split among equations.
 

J77

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What a bizarre problem to create.

I like dextor's answer.

Still loks like homework to me... :tongue:
 
360
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What a bizarre problem to create.

I like dextor's answer.
Gee J77...
I would rather like to call me a sick b*****d and to try to do something like Gib Z,than to recognize dextor's most creative answer as the mathematically valid solution.
:smile:
 

AlephZero

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2)What is given is the only positive solution to it,and what should one notice is that there is certain level of symmetry of 3 variable function split among equations.
I see the symmetry argument (and it's hard to write anything more about that without giving the complete answer, so I won't)

What I don't see is why you are restricting the solutions to positive quantities only. Why does that matter?
 

Gib Z

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haha techno, you love playing with peoples heads don't ya. You know when you said to dextor, the thing about computers solving quadratics, I thought that was a hint lol, i tried to reduce it to a quadratic for hours...thanks for telling me it was hopeless :D
 
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this is probably some kinda of olympiad-like problem. you probably need AM-GM, Cauchy inequalities or something like that.

from the first equation, you can easily show that 1/8>=xyz, somehow you need to show that (from the other equations):
xyz>=1/8 and the problem will be solved.

I'm not that fluid in these kinds of problem solving... I have barely done any problem in inequality besides reading the proofs for them... probably one of the reasons why I suck at Putnam...

perhaps one can approach this problem using a different method... consider, the Lagrange multiplier problem: extremizing the function (in positive reals):
[tex]f=xy^3+yz^3+zx^3[/tex]
with the constraint:
[tex]C=x+y+z=3/2 [/tex]

and with the Lagrange multiplier being [itex]u[/itex]

or you can consider assuming x>=y and get some contradiction.
 
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360
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tim_lou is on the right track.That's the best thinking so far!
One needs calculus and deeper insight to the problem.
That's what I expect of solvers here.
 

Gib Z

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O jesus Im not up to that kinda math yet lol.
 
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What kind of math you are not up to?Having learned the basics of multivariable calculus , by rewritting the equations a little,the proof of the claim can be presented in ~3 lines.
Another aspect of the problem is that it requires certain doseage of creativity.Unfortunately ,only one poster of this thread presented his creative side and gave pointers in the right direction...
FYI,I created this "homework " problem exploring the applications of:
http://cepa.newschool.edu/het/essays/theorem/euler.htm [Broken]

And I will not be giving any more hints than that. :smile:
 
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Using Lagrange multiplier... I bet one can minimize abc and show that the minimum is 1/2. but damn, that's like a 3 constraints, one function Lagrange multiplier problem... looks kinda messy.

by the way, I've just finished Calc three and have no idea what the heck Euler's Theorem is. looking from the link, it seems like it has something to do with vector space and eigenvalues and eigenvectors.... hmmm, linear Algebra.

I wonder if there is any elementary way of solving the system.
 
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Hurkyl

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There is a purely algoritmic way to arrive at the answer. You can systematically eliminate variables until, say, x is the only one remaining: at which point you'll have a single polynomial equation in x which has only one positive root.

So, you substitute x = 1/2 and repeat the process.

In fact, I see how to do the elimination quickly:
Use the second equation to eliminate u from the others.
Use the first equation to eliminate z from the last two.
Compute a resultant of the remaining equations to eliminate y.
Voila! What remains is a polynomial in x.
 
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that would probably require the use of cubic formula... with is ugly and inelegant.

edit: hmm maybe not, knowing that 1/2 is a solution, things should be reduced to quadratic...
 

Hurkyl

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I wasn't even thinking that far ahead: you can eliminate variables simply by adding multiples of one equation to other equations.

I think I'm going to be lazy, though, and not bother with grinding through the algebra. I don't want to compute (3 - 2x - 2y)^3.
 

Gib Z

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I won't bother either, but since I already did (x+y+z)^3, Ill post that and someone can substitute in for us.

[tex](x+y+z)^3=x^3+3xy^2+3x^2y+3x^2z+6xyz+y^3+3y^2z+z^3+3xz^2+3yz^2[/tex]
 
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In fact, I see how to do the elimination quickly:
Use the second equation to eliminate u from the others.
Use the first equation to eliminate z from the last two.
Compute a resultant of the remaining equations to eliminate y.
Voila! What remains is a polynomial in x.
2.>>> [tex]u=x^3+3yz^2[/tex]



1. >>>[tex]z=-x-y+3/2[/tex]

Using this,from 3th and 4th equation we have:
[tex]y^3+3x^2z-x^3-3yz^2=0;z^3+3xy^2-x^3-3yz^2=0[/tex]

Than how to proceed? Computing the resultant of these two equations for the variable y (the first one is cubic and the second is a quadratic) one eliminates y .
But here is " the but" again:
After introducing z=-x-y-3/2, I'm not arriving at cubic polynomial in variable x there.
Equating results for y you get expression containing only single variable x yes,but it looks horrible.
Anycase it's worse than cubic polynomial .
:frown:
 
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Hurkyl

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It sounds like you did it in the wrong order: eliminate u, then eliminate z, to get 2 cubics in x and y. The resultant should give you a nonic in x, which will have at least a double root at x = 1/2 because of the singularity of the system of equations (I expect a triple root because of the symmetry of the equations). I suspect it won't have any more positive roots -- if it does, then there is some other work to do. :frown:
 
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It sounds like you did it in the wrong order: eliminate u, then eliminate z, to get 2 cubics in x and y. The resultant should give you a nonic in x, which will have at least a double root at x = 1/2 because of the singularity of the system of equations (I expect a triple root because of the symmetry of the equations).
Wrong order or not to correct myself :
From 3th and 4th you indeed get the system of 2 cubics in x,y.But that can't be better ,in this case, than the system of one quadratic and one cubic.
Of course,there is Cardano formulae for cubics.Say we solve in y these 2 cubics.
But ,when we equate two solutions of the cubics don't we get the equation in x which is ,after rearangment and killing all the roots, of higher order than cubic?Unsolvable.That was the point.
However,there is still Sturm's theorem dealing with separtation of real roots from the complex ones of any single polynomial variable (any degree),and we know at least one real root here x=1/2 .Hence we can get away this time (in extremly ugly way though).
But what would be with your receipe if I asked for the proof that (1,1,1,6) is the only positive real solution of the system:

[tex]x+y+z-3=0[/tex]
[tex]x^5+5yz^4-u=0[/tex]
[tex]y^5+5x^4z-u=0[/tex]
[tex]z^5+5xy^4-u=0[/tex]

:confused:
Hehehe.
 

Gib Z

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Sometimes Our Solutions are inelegant and ugly, but far simpler than an elegant solution. The Proof for Fermat's Last theorem I heard is extremely ugly, but find us another one?...Exactly
 
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Gib Z,I got you new system _^^_ with exactly same type symmetry structure.Can you find an extremly ugly or elegant math proof now?:smile:

Sometimes Our Solutions are inelegant and ugly, but far simpler than an elegant solution. The Proof for Fermat's Last theorem I heard is extremely ugly, but find us another one?...Exactly
I don't understand ugly solution for Fermat's Last theorem so I can't judge wether it is ugly or nice.:smile:
Theory of numbers got nothing to do with my problem.
 

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