Number of real Solution [on hold]












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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







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put on hold as off-topic by José Carlos Santos, Saad, max_zorn, RRL, Javi Mar 29 at 22:45


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    -1












    $begingroup$


    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











    $endgroup$



    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.



















      -1












      -1








      -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











      $endgroup$




      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






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      share|cite|improve this question













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      edited Mar 29 at 17:04









      user1952500

      841712




      841712










      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.






















          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
          $$
          x=frac{ - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8}{356},
          $$



          $$
          y=frac{6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139}{712},
          $$
          where $z$ is a (non-real) complex root of the polynomial
          $$
          89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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





















          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 $f(t) = dfrac{2t^2}{1+t^2}, t in mathbb{R}^{+}$. 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 !






          share|cite|improve this answer









          $endgroup$




















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            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
            $$
            x=frac{ - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8}{356},
            $$



            $$
            y=frac{6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139}{712},
            $$
            where $z$ is a (non-real) complex root of the polynomial
            $$
            89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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


















            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
            $$
            x=frac{ - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8}{356},
            $$



            $$
            y=frac{6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139}{712},
            $$
            where $z$ is a (non-real) complex root of the polynomial
            $$
            89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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
















            5












            5








            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
            $$
            x=frac{ - 445z^5 - 962z^4 - 110z^3 + 654z^2 - 21z - 8}{356},
            $$



            $$
            y=frac{6942z^5 + 9863z^4 + 1852z^3 + 902z^2 + 214z + 139}{712},
            $$
            where $z$ is a (non-real) complex root of the polynomial
            $$
            89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
            $$






            share|cite|improve this answer









            $endgroup$



            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
            $$
            89z^6 + 50z^5 + 31z^4 + 12z^3 + 7z^2 + 2z + 1.
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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




















            • $begingroup$
              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













            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 $f(t) = dfrac{2t^2}{1+t^2}, t in mathbb{R}^{+}$. 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 !






            share|cite|improve this answer









            $endgroup$


















              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 $f(t) = dfrac{2t^2}{1+t^2}, t in mathbb{R}^{+}$. 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 !






              share|cite|improve this answer









              $endgroup$
















                5












                5








                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 $f(t) = dfrac{2t^2}{1+t^2}, t in mathbb{R}^{+}$. 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 !






                share|cite|improve this answer









                $endgroup$



                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 !







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 29 at 17:15









                DeepSeaDeepSea

                71.4k54488




                71.4k54488















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