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Finding the roots of a 2 variable polynomial


Finding the sum of non-real roots of a polynomial.not complete polynomial and rootsPolynomial with real rootsDetermine polynomial whose roots are a linear combination of roots of another polynomialPolynomial with odd number of real rootsPolynomial Root Finding Algorithmfinding roots and factors of a polynomialroots of the polynomialHow to find out how many roots of a polynomial is in each quadrant?Necessary and sufficient conditions for the existence of a focus in an ODE













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$begingroup$


Consider the polynomial $x^{n+1}+y^{n+1}+yx^n$ where $ninmathbb{N}$ is odd. I would like to find the real roots of this polynomial. I believe the only root is $(x,y)=(0,0)$. So far, I've been able to show that if we have a root different than $(0,0)$ then either $x>0,y<0$ or $x<0,y>0$. One would only have to check for one of the cases, since the polynomial is homogeneous. However, I'm completely stuck. Any ideas? Thank you in advance.






polynomials homogeneous-equation







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Consider the polynomial $x^{n+1}+y^{n+1}+yx^n$ where $ninmathbb{N}$ is odd. I would like to find the real roots of this polynomial. I believe the only root is $(x,y)=(0,0)$. So far, I've been able to show that if we have a root different than $(0,0)$ then either $x>0,y<0$ or $x<0,y>0$. One would only have to check for one of the cases, since the polynomial is homogeneous. However, I'm completely stuck. Any ideas? Thank you in advance.






    polynomials homogeneous-equation







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Consider the polynomial $x^{n+1}+y^{n+1}+yx^n$ where $ninmathbb{N}$ is odd. I would like to find the real roots of this polynomial. I believe the only root is $(x,y)=(0,0)$. So far, I've been able to show that if we have a root different than $(0,0)$ then either $x>0,y<0$ or $x<0,y>0$. One would only have to check for one of the cases, since the polynomial is homogeneous. However, I'm completely stuck. Any ideas? Thank you in advance.






      polynomials homogeneous-equation







      share|cite|improve this question









      $endgroup$




      Consider the polynomial $x^{n+1}+y^{n+1}+yx^n$ where $ninmathbb{N}$ is odd. I would like to find the real roots of this polynomial. I believe the only root is $(x,y)=(0,0)$. So far, I've been able to show that if we have a root different than $(0,0)$ then either $x>0,y<0$ or $x<0,y>0$. One would only have to check for one of the cases, since the polynomial is homogeneous. However, I'm completely stuck. Any ideas? Thank you in advance.






      polynomials homogeneous-equation




      polynomials homogeneous-equation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 9 at 17:03









      Ray BernRay Bern

      15813




      15813






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Using the well-known inequality $ab le frac{a^p}{p} + frac{b^q}{q}$ for $a, , b ge 0$, where $p^{-1} + q^{-1} = 1$, one can estimate
          $$
          yx^n ge - |y||x|^n ge - frac{|y|^{n+1}}{n+1} - frac{n|x|^{n+1}}{n+1}
          $$
          and therefore, since $n+1$ is even,
          $$
          y^{n+1} +y^{n+1} + yx^n ge frac{n}{n+1}y^{n+1} + frac{1}{n+1}x^{n+1} , .
          $$
          This is positive except if $x = y = 0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wow, did not expect to take this path, thank you!
            $endgroup$
            – Ray Bern
            Mar 9 at 17:19



















          0












          $begingroup$

          As the polynomial is homogeneous, you solve



          $$t^{n+1}+t+1=0$$ and all $(x,tx)$ are solutions.



          But it doesn't take long to see that the polynomial has no real roots.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Using the well-known inequality $ab le frac{a^p}{p} + frac{b^q}{q}$ for $a, , b ge 0$, where $p^{-1} + q^{-1} = 1$, one can estimate
            $$
            yx^n ge - |y||x|^n ge - frac{|y|^{n+1}}{n+1} - frac{n|x|^{n+1}}{n+1}
            $$
            and therefore, since $n+1$ is even,
            $$
            y^{n+1} +y^{n+1} + yx^n ge frac{n}{n+1}y^{n+1} + frac{1}{n+1}x^{n+1} , .
            $$
            This is positive except if $x = y = 0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Wow, did not expect to take this path, thank you!
              $endgroup$
              – Ray Bern
              Mar 9 at 17:19
















            1












            $begingroup$

            Using the well-known inequality $ab le frac{a^p}{p} + frac{b^q}{q}$ for $a, , b ge 0$, where $p^{-1} + q^{-1} = 1$, one can estimate
            $$
            yx^n ge - |y||x|^n ge - frac{|y|^{n+1}}{n+1} - frac{n|x|^{n+1}}{n+1}
            $$
            and therefore, since $n+1$ is even,
            $$
            y^{n+1} +y^{n+1} + yx^n ge frac{n}{n+1}y^{n+1} + frac{1}{n+1}x^{n+1} , .
            $$
            This is positive except if $x = y = 0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Wow, did not expect to take this path, thank you!
              $endgroup$
              – Ray Bern
              Mar 9 at 17:19














            1












            1








            1





            $begingroup$

            Using the well-known inequality $ab le frac{a^p}{p} + frac{b^q}{q}$ for $a, , b ge 0$, where $p^{-1} + q^{-1} = 1$, one can estimate
            $$
            yx^n ge - |y||x|^n ge - frac{|y|^{n+1}}{n+1} - frac{n|x|^{n+1}}{n+1}
            $$
            and therefore, since $n+1$ is even,
            $$
            y^{n+1} +y^{n+1} + yx^n ge frac{n}{n+1}y^{n+1} + frac{1}{n+1}x^{n+1} , .
            $$
            This is positive except if $x = y = 0$.






            share|cite|improve this answer











            $endgroup$



            Using the well-known inequality $ab le frac{a^p}{p} + frac{b^q}{q}$ for $a, , b ge 0$, where $p^{-1} + q^{-1} = 1$, one can estimate
            $$
            yx^n ge - |y||x|^n ge - frac{|y|^{n+1}}{n+1} - frac{n|x|^{n+1}}{n+1}
            $$
            and therefore, since $n+1$ is even,
            $$
            y^{n+1} +y^{n+1} + yx^n ge frac{n}{n+1}y^{n+1} + frac{1}{n+1}x^{n+1} , .
            $$
            This is positive except if $x = y = 0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 9 at 17:40

























            answered Mar 9 at 17:10









            Hans EnglerHans Engler

            10.5k11836




            10.5k11836












            • $begingroup$
              Wow, did not expect to take this path, thank you!
              $endgroup$
              – Ray Bern
              Mar 9 at 17:19


















            • $begingroup$
              Wow, did not expect to take this path, thank you!
              $endgroup$
              – Ray Bern
              Mar 9 at 17:19
















            $begingroup$
            Wow, did not expect to take this path, thank you!
            $endgroup$
            – Ray Bern
            Mar 9 at 17:19




            $begingroup$
            Wow, did not expect to take this path, thank you!
            $endgroup$
            – Ray Bern
            Mar 9 at 17:19











            0












            $begingroup$

            As the polynomial is homogeneous, you solve



            $$t^{n+1}+t+1=0$$ and all $(x,tx)$ are solutions.



            But it doesn't take long to see that the polynomial has no real roots.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              As the polynomial is homogeneous, you solve



              $$t^{n+1}+t+1=0$$ and all $(x,tx)$ are solutions.



              But it doesn't take long to see that the polynomial has no real roots.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                As the polynomial is homogeneous, you solve



                $$t^{n+1}+t+1=0$$ and all $(x,tx)$ are solutions.



                But it doesn't take long to see that the polynomial has no real roots.






                share|cite|improve this answer









                $endgroup$



                As the polynomial is homogeneous, you solve



                $$t^{n+1}+t+1=0$$ and all $(x,tx)$ are solutions.



                But it doesn't take long to see that the polynomial has no real roots.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 9 at 17:19









                Yves DaoustYves Daoust

                130k676227




                130k676227






























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