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Number of real Solution [on hold]

-1

2

$begingroup$

Find the number of real Solution of the system of equations

$$ x = frac 2z^2 1+z^2 , y = frac 2x^2 1+x^2 , z = frac 2y^2 1+y^2 $$

systems-of-equations

share|cite|improve this question

edited 2 days ago

user1952500

841712

asked 2 days ago

user157835user157835

10417

$endgroup$

put on hold as off-topic by José Carlos Santos, Saad, max_zorn, RRL, Javi 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Saad, max_zorn, RRL, Javi

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

-1

2

$begingroup$

Find the number of real Solution of the system of equations

$$ x = frac 2z^2 1+z^2 , y = frac 2x^2 1+x^2 , z = frac 2y^2 1+y^2 $$

systems-of-equations

share|cite|improve this question

edited 2 days ago

user1952500

841712

asked 2 days ago

user157835user157835

10417

$endgroup$

put on hold as off-topic by José Carlos Santos, Saad, max_zorn, RRL, Javi 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Saad, max_zorn, RRL, Javi

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

$begingroup$

Find the number of real Solution of the system of equations

$$ x = frac 2z^2 1+z^2 , y = frac 2x^2 1+x^2 , z = frac 2y^2 1+y^2 $$

systems-of-equations

share|cite|improve this question

edited 2 days ago

user1952500

841712

asked 2 days ago

user157835user157835

10417

$endgroup$

share|cite|improve this question

edited 2 days ago

user1952500

841712

asked 2 days ago

user157835user157835

10417

share|cite|improve this question

edited 2 days ago

user1952500

841712

asked 2 days ago

user157835user157835

10417

edited 2 days ago

user1952500

841712

edited 2 days ago

user1952500

841712

asked 2 days ago

user157835user157835

10417

asked 2 days ago

user157835user157835

10417

2 Answers
2

active

oldest

votes

5

$begingroup$

We have the obvious solutions $(x,y,z)=(0,0,0)=(1,1,1)$. Over the real numbers these are all, but over the complex numbers there are six additional solutions, given by
$$
x=frac - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8356,
$$

$$
y=frac6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139712,
$$
where $z$ is a (non-real) complex root of the polynomial
$$
89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
$$

share|cite|improve this answer

answered 2 days ago

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, is there any approach to show that there will be no real Solution other than 1 and 0
  $endgroup$
  – user157835
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, there are several approaches. Solving the system with resultants gives all complex solutions. Cancelling out the non-real ones gives the real solutions.
  $endgroup$
  – Dietrich Burde
  2 days ago
  
  

  
  
  
  

add a comment | 

5

$begingroup$

The problem is not new and has appeared over here quite a few times as far as I could remember. Seeking the "real solutions" is perhaps more interesting than the "imaginary solutions". For the real solutions, a functional approach is my favorite. So let $f(t) = dfrac2t^21+t^2, t in mathbbR^+$. We have $f'(t) = dfrac4t(1+t^2)^2> 0$. Thus let's say if $x ge y ge z implies f(x) ge f(y) ge f(z)implies y ge z ge x implies x ge y ge z ge ximplies x = y = zimplies x = dfrac2x^21+x^2implies x = 0,1implies (x,y,z) = (0,0,0) ; (1,1,1)$,and these are all the (real) solutions !

share|cite|improve this answer

answered 2 days ago

DeepSeaDeepSea

71.4k54488

$endgroup$

add a comment | 

5

$begingroup$

We have the obvious solutions $(x,y,z)=(0,0,0)=(1,1,1)$. Over the real numbers these are all, but over the complex numbers there are six additional solutions, given by
$$
x=frac - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8356,
$$

$$
y=frac6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139712,
$$
where $z$ is a (non-real) complex root of the polynomial
$$
89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
$$

share|cite|improve this answer

answered 2 days ago

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, is there any approach to show that there will be no real Solution other than 1 and 0
  $endgroup$
  – user157835
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, there are several approaches. Solving the system with resultants gives all complex solutions. Cancelling out the non-real ones gives the real solutions.
  $endgroup$
  – Dietrich Burde
  2 days ago
  
  

  
  
  
  

add a comment | 

5

$begingroup$

We have the obvious solutions $(x,y,z)=(0,0,0)=(1,1,1)$. Over the real numbers these are all, but over the complex numbers there are six additional solutions, given by
$$
x=frac - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8356,
$$

$$
y=frac6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139712,
$$
where $z$ is a (non-real) complex root of the polynomial
$$
89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
$$

share|cite|improve this answer

answered 2 days ago

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, is there any approach to show that there will be no real Solution other than 1 and 0
  $endgroup$
  – user157835
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, there are several approaches. Solving the system with resultants gives all complex solutions. Cancelling out the non-real ones gives the real solutions.
  $endgroup$
  – Dietrich Burde
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

We have the obvious solutions $(x,y,z)=(0,0,0)=(1,1,1)$. Over the real numbers these are all, but over the complex numbers there are six additional solutions, given by
$$
x=frac - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8356,
$$

$$
y=frac6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139712,
$$
where $z$ is a (non-real) complex root of the polynomial
$$
89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
$$

share|cite|improve this answer

answered 2 days ago

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

share|cite|improve this answer

answered 2 days ago

Dietrich BurdeDietrich Burde

81.6k648106

answered 2 days ago

Dietrich BurdeDietrich Burde

81.6k648106

answered 2 days ago

Dietrich BurdeDietrich Burde

81.6k648106

5

$begingroup$

The problem is not new and has appeared over here quite a few times as far as I could remember. Seeking the "real solutions" is perhaps more interesting than the "imaginary solutions". For the real solutions, a functional approach is my favorite. So let $f(t) = dfrac2t^21+t^2, t in mathbbR^+$. We have $f'(t) = dfrac4t(1+t^2)^2> 0$. Thus let's say if $x ge y ge z implies f(x) ge f(y) ge f(z)implies y ge z ge x implies x ge y ge z ge ximplies x = y = zimplies x = dfrac2x^21+x^2implies x = 0,1implies (x,y,z) = (0,0,0) ; (1,1,1)$,and these are all the (real) solutions !

share|cite|improve this answer

answered 2 days ago

DeepSeaDeepSea

71.4k54488

$endgroup$

add a comment | 

5

$begingroup$

The problem is not new and has appeared over here quite a few times as far as I could remember. Seeking the "real solutions" is perhaps more interesting than the "imaginary solutions". For the real solutions, a functional approach is my favorite. So let $f(t) = dfrac2t^21+t^2, t in mathbbR^+$. We have $f'(t) = dfrac4t(1+t^2)^2> 0$. Thus let's say if $x ge y ge z implies f(x) ge f(y) ge f(z)implies y ge z ge x implies x ge y ge z ge ximplies x = y = zimplies x = dfrac2x^21+x^2implies x = 0,1implies (x,y,z) = (0,0,0) ; (1,1,1)$,and these are all the (real) solutions !

share|cite|improve this answer

answered 2 days ago

DeepSeaDeepSea

71.4k54488

$endgroup$

add a comment | 

$begingroup$

The problem is not new and has appeared over here quite a few times as far as I could remember. Seeking the "real solutions" is perhaps more interesting than the "imaginary solutions". For the real solutions, a functional approach is my favorite. So let $f(t) = dfrac2t^21+t^2, t in mathbbR^+$. We have $f'(t) = dfrac4t(1+t^2)^2> 0$. Thus let's say if $x ge y ge z implies f(x) ge f(y) ge f(z)implies y ge z ge x implies x ge y ge z ge ximplies x = y = zimplies x = dfrac2x^21+x^2implies x = 0,1implies (x,y,z) = (0,0,0) ; (1,1,1)$,and these are all the (real) solutions !

share|cite|improve this answer

answered 2 days ago

DeepSeaDeepSea

71.4k54488

$endgroup$

share|cite|improve this answer

answered 2 days ago

DeepSeaDeepSea

71.4k54488

answered 2 days ago

DeepSeaDeepSea

71.4k54488

answered 2 days ago

DeepSeaDeepSea

71.4k54488