Number of real Solution [on hold]
$begingroup$
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
edited Mar 29 at 17:04
user1952500
841712
asked Mar 29 at 16:31
user157835user157835
10417
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put on hold as off-topic by José Carlos Santos, Saad, max_zorn, RRL, Javi Mar 29 at 22:45
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
edited Mar 29 at 17:04
user1952500
841712
asked Mar 29 at 16:31
user157835user157835
10417
$endgroup$
put on hold as off-topic by José Carlos Santos, Saad, max_zorn, RRL, Javi Mar 29 at 22:45
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
edited Mar 29 at 17:04
user1952500
841712
asked Mar 29 at 16:31
user157835user157835
10417
$endgroup$
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
systems-of-equations
edited Mar 29 at 17:04
user1952500
841712
asked Mar 29 at 16:31
user157835user157835
10417
edited Mar 29 at 17:04
user1952500
841712
asked Mar 29 at 16:31
user157835user157835
10417
edited Mar 29 at 17:04
user1952500
841712
edited Mar 29 at 17:04
user1952500
841712
edited Mar 29 at 17:04
user1952500
841712
841712
asked Mar 29 at 16:31
user157835user157835
10417
asked Mar 29 at 16:31
user157835user157835
10417
asked Mar 29 at 16:31
user157835user157835
10417
10417
put on hold as off-topic by José Carlos Santos, Saad, max_zorn, RRL, Javi Mar 29 at 22:45
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.
put on hold as off-topic by José Carlos Santos, Saad, max_zorn, RRL, Javi Mar 29 at 22:45
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 |
add a comment |
2 Answers
2
active
oldest
votes
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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
$$
x=frac{ - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8}{356},
$$
$$
y=frac{6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139}{712},
$$
where $z$ is a (non-real) complex root of the polynomial
$$
89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
$$
answered Mar 29 at 16:39
Dietrich BurdeDietrich Burde
81.5k648106
$endgroup$
$begingroup$
Thanks, is there any approach to show that there will be no real Solution other than 1 and 0
$endgroup$
– user157835
Mar 29 at 16:47
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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
Mar 29 at 19:32
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 $f(t) = dfrac{2t^2}{1+t^2}, t in mathbb{R}^{+}$. We have $f'(t) = dfrac{4t}{(1+t^2)^2}> 0$. Thus let's say if $x ge y ge z implies f(x) ge f(y) ge f(z)implies y ge z ge x implies x ge y ge z ge ximplies x = y = zimplies x = dfrac{2x^2}{1+x^2}implies x = 0,1implies (x,y,z) = (0,0,0) ; (1,1,1)$,and these are all the (real) solutions !
answered Mar 29 at 17:15
DeepSeaDeepSea
71.4k54488
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$
x=frac{ - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8}{356},
$$
$$
y=frac{6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139}{712},
$$
where $z$ is a (non-real) complex root of the polynomial
$$
89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
$$
answered Mar 29 at 16:39
Dietrich BurdeDietrich Burde
81.5k648106
$endgroup$
$begingroup$
Thanks, is there any approach to show that there will be no real Solution other than 1 and 0
$endgroup$
– user157835
Mar 29 at 16:47
$begingroup$
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
Mar 29 at 19:32
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
$$
x=frac{ - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8}{356},
$$
$$
y=frac{6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139}{712},
$$
where $z$ is a (non-real) complex root of the polynomial
$$
89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
$$
answered Mar 29 at 16:39
Dietrich BurdeDietrich Burde
81.5k648106
$endgroup$
$begingroup$
Thanks, is there any approach to show that there will be no real Solution other than 1 and 0
$endgroup$
– user157835
Mar 29 at 16:47
$begingroup$
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
Mar 29 at 19:32
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
$$
x=frac{ - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8}{356},
$$
$$
y=frac{6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139}{712},
$$
where $z$ is a (non-real) complex root of the polynomial
$$
89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
$$
answered Mar 29 at 16:39
Dietrich BurdeDietrich Burde
81.5k648106
$endgroup$
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
$$
x=frac{ - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8}{356},
$$
$$
y=frac{6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139}{712},
$$
where $z$ is a (non-real) complex root of the polynomial
$$
89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
$$
answered Mar 29 at 16:39
Dietrich BurdeDietrich Burde
81.5k648106
answered Mar 29 at 16:39
Dietrich BurdeDietrich Burde
81.5k648106
answered Mar 29 at 16:39
Dietrich BurdeDietrich Burde
81.5k648106
answered Mar 29 at 16:39
Dietrich BurdeDietrich Burde
81.5k648106
81.5k648106
$begingroup$
Thanks, is there any approach to show that there will be no real Solution other than 1 and 0
$endgroup$
– user157835
Mar 29 at 16:47
$begingroup$
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
Mar 29 at 19:32
add a comment |
$begingroup$
Thanks, is there any approach to show that there will be no real Solution other than 1 and 0
$endgroup$
– user157835
Mar 29 at 16:47
$begingroup$
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
Mar 29 at 19:32
$begingroup$
Thanks, is there any approach to show that there will be no real Solution other than 1 and 0
$endgroup$
– user157835
Mar 29 at 16:47
$begingroup$
Thanks, is there any approach to show that there will be no real Solution other than 1 and 0
$endgroup$
– user157835
Mar 29 at 16:47
$begingroup$
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
Mar 29 at 19:32
$begingroup$
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
Mar 29 at 19:32
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 $f(t) = dfrac{2t^2}{1+t^2}, t in mathbb{R}^{+}$. We have $f'(t) = dfrac{4t}{(1+t^2)^2}> 0$. Thus let's say if $x ge y ge z implies f(x) ge f(y) ge f(z)implies y ge z ge x implies x ge y ge z ge ximplies x = y = zimplies x = dfrac{2x^2}{1+x^2}implies x = 0,1implies (x,y,z) = (0,0,0) ; (1,1,1)$,and these are all the (real) solutions !
answered Mar 29 at 17:15
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 $f(t) = dfrac{2t^2}{1+t^2}, t in mathbb{R}^{+}$. We have $f'(t) = dfrac{4t}{(1+t^2)^2}> 0$. Thus let's say if $x ge y ge z implies f(x) ge f(y) ge f(z)implies y ge z ge x implies x ge y ge z ge ximplies x = y = zimplies x = dfrac{2x^2}{1+x^2}implies x = 0,1implies (x,y,z) = (0,0,0) ; (1,1,1)$,and these are all the (real) solutions !
answered Mar 29 at 17:15
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 $f(t) = dfrac{2t^2}{1+t^2}, t in mathbb{R}^{+}$. We have $f'(t) = dfrac{4t}{(1+t^2)^2}> 0$. Thus let's say if $x ge y ge z implies f(x) ge f(y) ge f(z)implies y ge z ge x implies x ge y ge z ge ximplies x = y = zimplies x = dfrac{2x^2}{1+x^2}implies x = 0,1implies (x,y,z) = (0,0,0) ; (1,1,1)$,and these are all the (real) solutions !
answered Mar 29 at 17:15
DeepSeaDeepSea
71.4k54488
$endgroup$
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 $f(t) = dfrac{2t^2}{1+t^2}, t in mathbb{R}^{+}$. We have $f'(t) = dfrac{4t}{(1+t^2)^2}> 0$. Thus let's say if $x ge y ge z implies f(x) ge f(y) ge f(z)implies y ge z ge x implies x ge y ge z ge ximplies x = y = zimplies x = dfrac{2x^2}{1+x^2}implies x = 0,1implies (x,y,z) = (0,0,0) ; (1,1,1)$,and these are all the (real) solutions !
answered Mar 29 at 17:15
DeepSeaDeepSea
71.4k54488
answered Mar 29 at 17:15
DeepSeaDeepSea
71.4k54488
answered Mar 29 at 17:15
DeepSeaDeepSea
71.4k54488
answered Mar 29 at 17:15
DeepSeaDeepSea
71.4k54488
71.4k54488
add a comment |
add a comment |
