Find critical points (minima) from gradient (2D vector)

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I'm trying to find critical points (minima) for a gradient in 2D space. The following are partial derivative and its polynomial equation:
enter image description here



partial derivative



To find critical points I will need to set gradient to 0 and solve for 2 vectors span across $x_1$ and $x_2$. Could anyone provide steps in finding critical point for multi-variable equations? I'm trying to self-study and having this problem solved as a reference will be very helpful to my learning progress.



Thank you!




linear-algebra systems-of-equations maxima-minima




share|cite|improve this question

















  • 2




    The main problem is the solution for a nonlinear equations system. In general this should be done using iterative numerical methods but in the present case, due to a special formula structure, the solution is quite simple. Think that adding the two equations you get $8(x_1+x_2) = 0$ so $x_1 = - x_2$ and now the solution appears $-4x_1^3+8 x_1+1=0$ that can be solved by Cardano's formula.
    – Cesareo
    Jul 22 at 8:31










  • The objective function is polynomial. Hence, so is the gradient. Thus, finding where the gradient vanishes boils down to solving a system of polynomial equations. In this case, it is easy. For harder systems, take a look at Gröbner bases.
    – Rodrigo de Azevedo
    Jul 22 at 15:22











  • @RodrigodeAzevedo My initial thought of the question was not about the concept , but more of how I’m able solve the problem in exam setting with limited time available. Is Cardano the only way to solve for minima vectors?
    – lydias
    Jul 22 at 15:28










  • If you add the two cubic equations, you obtain $8 x_1 + 8 x_2 = 0$. Hence, $x_2 = -x_1$, or $x_1 = - x_2$. Substituting, you obtain a cubic equation in either $x_1$ or $x_2$. Some cubic equations can be solved by visual inspection, if the exam question is designed properly. I do not know what you mean by "Cardano".
    – Rodrigo de Azevedo
    Jul 22 at 15:33











  • @RodrigodeAzevedo Cardano as in Cardano formula, which will be very time consuming in exam setting.
    – lydias
    Jul 22 at 15:35














up vote
0
down vote

favorite












I'm trying to find critical points (minima) for a gradient in 2D space. The following are partial derivative and its polynomial equation:
enter image description here



partial derivative



To find critical points I will need to set gradient to 0 and solve for 2 vectors span across $x_1$ and $x_2$. Could anyone provide steps in finding critical point for multi-variable equations? I'm trying to self-study and having this problem solved as a reference will be very helpful to my learning progress.



Thank you!




linear-algebra systems-of-equations maxima-minima




share|cite|improve this question

















  • 2




    The main problem is the solution for a nonlinear equations system. In general this should be done using iterative numerical methods but in the present case, due to a special formula structure, the solution is quite simple. Think that adding the two equations you get $8(x_1+x_2) = 0$ so $x_1 = - x_2$ and now the solution appears $-4x_1^3+8 x_1+1=0$ that can be solved by Cardano's formula.
    – Cesareo
    Jul 22 at 8:31










  • The objective function is polynomial. Hence, so is the gradient. Thus, finding where the gradient vanishes boils down to solving a system of polynomial equations. In this case, it is easy. For harder systems, take a look at Gröbner bases.
    – Rodrigo de Azevedo
    Jul 22 at 15:22











  • @RodrigodeAzevedo My initial thought of the question was not about the concept , but more of how I’m able solve the problem in exam setting with limited time available. Is Cardano the only way to solve for minima vectors?
    – lydias
    Jul 22 at 15:28










  • If you add the two cubic equations, you obtain $8 x_1 + 8 x_2 = 0$. Hence, $x_2 = -x_1$, or $x_1 = - x_2$. Substituting, you obtain a cubic equation in either $x_1$ or $x_2$. Some cubic equations can be solved by visual inspection, if the exam question is designed properly. I do not know what you mean by "Cardano".
    – Rodrigo de Azevedo
    Jul 22 at 15:33











  • @RodrigodeAzevedo Cardano as in Cardano formula, which will be very time consuming in exam setting.
    – lydias
    Jul 22 at 15:35












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm trying to find critical points (minima) for a gradient in 2D space. The following are partial derivative and its polynomial equation:
enter image description here



partial derivative



To find critical points I will need to set gradient to 0 and solve for 2 vectors span across $x_1$ and $x_2$. Could anyone provide steps in finding critical point for multi-variable equations? I'm trying to self-study and having this problem solved as a reference will be very helpful to my learning progress.



Thank you!




linear-algebra systems-of-equations maxima-minima




share|cite|improve this question













I'm trying to find critical points (minima) for a gradient in 2D space. The following are partial derivative and its polynomial equation:
enter image description here



partial derivative



To find critical points I will need to set gradient to 0 and solve for 2 vectors span across $x_1$ and $x_2$. Could anyone provide steps in finding critical point for multi-variable equations? I'm trying to self-study and having this problem solved as a reference will be very helpful to my learning progress.



Thank you!





linear-algebra systems-of-equations maxima-minima





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 22 at 15:29
























asked Jul 22 at 4:07









lydias

808




808







  • 2




    The main problem is the solution for a nonlinear equations system. In general this should be done using iterative numerical methods but in the present case, due to a special formula structure, the solution is quite simple. Think that adding the two equations you get $8(x_1+x_2) = 0$ so $x_1 = - x_2$ and now the solution appears $-4x_1^3+8 x_1+1=0$ that can be solved by Cardano's formula.
    – Cesareo
    Jul 22 at 8:31










  • The objective function is polynomial. Hence, so is the gradient. Thus, finding where the gradient vanishes boils down to solving a system of polynomial equations. In this case, it is easy. For harder systems, take a look at Gröbner bases.
    – Rodrigo de Azevedo
    Jul 22 at 15:22











  • @RodrigodeAzevedo My initial thought of the question was not about the concept , but more of how I’m able solve the problem in exam setting with limited time available. Is Cardano the only way to solve for minima vectors?
    – lydias
    Jul 22 at 15:28










  • If you add the two cubic equations, you obtain $8 x_1 + 8 x_2 = 0$. Hence, $x_2 = -x_1$, or $x_1 = - x_2$. Substituting, you obtain a cubic equation in either $x_1$ or $x_2$. Some cubic equations can be solved by visual inspection, if the exam question is designed properly. I do not know what you mean by "Cardano".
    – Rodrigo de Azevedo
    Jul 22 at 15:33











  • @RodrigodeAzevedo Cardano as in Cardano formula, which will be very time consuming in exam setting.
    – lydias
    Jul 22 at 15:35












  • 2




    The main problem is the solution for a nonlinear equations system. In general this should be done using iterative numerical methods but in the present case, due to a special formula structure, the solution is quite simple. Think that adding the two equations you get $8(x_1+x_2) = 0$ so $x_1 = - x_2$ and now the solution appears $-4x_1^3+8 x_1+1=0$ that can be solved by Cardano's formula.
    – Cesareo
    Jul 22 at 8:31










  • The objective function is polynomial. Hence, so is the gradient. Thus, finding where the gradient vanishes boils down to solving a system of polynomial equations. In this case, it is easy. For harder systems, take a look at Gröbner bases.
    – Rodrigo de Azevedo
    Jul 22 at 15:22











  • @RodrigodeAzevedo My initial thought of the question was not about the concept , but more of how I’m able solve the problem in exam setting with limited time available. Is Cardano the only way to solve for minima vectors?
    – lydias
    Jul 22 at 15:28










  • If you add the two cubic equations, you obtain $8 x_1 + 8 x_2 = 0$. Hence, $x_2 = -x_1$, or $x_1 = - x_2$. Substituting, you obtain a cubic equation in either $x_1$ or $x_2$. Some cubic equations can be solved by visual inspection, if the exam question is designed properly. I do not know what you mean by "Cardano".
    – Rodrigo de Azevedo
    Jul 22 at 15:33











  • @RodrigodeAzevedo Cardano as in Cardano formula, which will be very time consuming in exam setting.
    – lydias
    Jul 22 at 15:35







2




2




The main problem is the solution for a nonlinear equations system. In general this should be done using iterative numerical methods but in the present case, due to a special formula structure, the solution is quite simple. Think that adding the two equations you get $8(x_1+x_2) = 0$ so $x_1 = - x_2$ and now the solution appears $-4x_1^3+8 x_1+1=0$ that can be solved by Cardano's formula.
– Cesareo
Jul 22 at 8:31




The main problem is the solution for a nonlinear equations system. In general this should be done using iterative numerical methods but in the present case, due to a special formula structure, the solution is quite simple. Think that adding the two equations you get $8(x_1+x_2) = 0$ so $x_1 = - x_2$ and now the solution appears $-4x_1^3+8 x_1+1=0$ that can be solved by Cardano's formula.
– Cesareo
Jul 22 at 8:31












The objective function is polynomial. Hence, so is the gradient. Thus, finding where the gradient vanishes boils down to solving a system of polynomial equations. In this case, it is easy. For harder systems, take a look at Gröbner bases.
– Rodrigo de Azevedo
Jul 22 at 15:22





The objective function is polynomial. Hence, so is the gradient. Thus, finding where the gradient vanishes boils down to solving a system of polynomial equations. In this case, it is easy. For harder systems, take a look at Gröbner bases.
– Rodrigo de Azevedo
Jul 22 at 15:22













@RodrigodeAzevedo My initial thought of the question was not about the concept , but more of how I’m able solve the problem in exam setting with limited time available. Is Cardano the only way to solve for minima vectors?
– lydias
Jul 22 at 15:28




@RodrigodeAzevedo My initial thought of the question was not about the concept , but more of how I’m able solve the problem in exam setting with limited time available. Is Cardano the only way to solve for minima vectors?
– lydias
Jul 22 at 15:28












If you add the two cubic equations, you obtain $8 x_1 + 8 x_2 = 0$. Hence, $x_2 = -x_1$, or $x_1 = - x_2$. Substituting, you obtain a cubic equation in either $x_1$ or $x_2$. Some cubic equations can be solved by visual inspection, if the exam question is designed properly. I do not know what you mean by "Cardano".
– Rodrigo de Azevedo
Jul 22 at 15:33





If you add the two cubic equations, you obtain $8 x_1 + 8 x_2 = 0$. Hence, $x_2 = -x_1$, or $x_1 = - x_2$. Substituting, you obtain a cubic equation in either $x_1$ or $x_2$. Some cubic equations can be solved by visual inspection, if the exam question is designed properly. I do not know what you mean by "Cardano".
– Rodrigo de Azevedo
Jul 22 at 15:33













@RodrigodeAzevedo Cardano as in Cardano formula, which will be very time consuming in exam setting.
– lydias
Jul 22 at 15:35




@RodrigodeAzevedo Cardano as in Cardano formula, which will be very time consuming in exam setting.
– lydias
Jul 22 at 15:35










1 Answer
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Setting the gradient of a multi variable equation equal to zero will give you the (local)minimum, (local)maximum or saddle points.



In your case, even though the gradient equation system is non-linear, we note that adding the two equations we get $$8x_1 + 8x_2 = 0 Rightarrow x_1=-x_2.$$



Now we just substitute it in either the first or the second equation of the gradient system. Doing it for the first equation yields
$$-4(x_2+x_2)^3+8x_2-1=0Rightarrow-32x_2^3+8x_2-1=0$$
Solving it (using numerical methods, for example) and using the fact that $x_1 = -x_2$, we get the following points:
$$(-0.13480, 0.13480), quad (-0.41878, 0.41878), quad (0.55358, -0.55358).$$



This is the contour plot of $F(x)$
enter image description here



Analyzing this plot (or doing it numerically) we conclude that:



  • $(-0.13480, 0.13480)$: saddle point;

  • $(-0.41878, 0.41878)$: local minimum;

  • $(0.55358, -0.55358)$: global minimum.





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    Setting the gradient of a multi variable equation equal to zero will give you the (local)minimum, (local)maximum or saddle points.



    In your case, even though the gradient equation system is non-linear, we note that adding the two equations we get $$8x_1 + 8x_2 = 0 Rightarrow x_1=-x_2.$$



    Now we just substitute it in either the first or the second equation of the gradient system. Doing it for the first equation yields
    $$-4(x_2+x_2)^3+8x_2-1=0Rightarrow-32x_2^3+8x_2-1=0$$
    Solving it (using numerical methods, for example) and using the fact that $x_1 = -x_2$, we get the following points:
    $$(-0.13480, 0.13480), quad (-0.41878, 0.41878), quad (0.55358, -0.55358).$$



    This is the contour plot of $F(x)$
    enter image description here



    Analyzing this plot (or doing it numerically) we conclude that:



    • $(-0.13480, 0.13480)$: saddle point;

    • $(-0.41878, 0.41878)$: local minimum;

    • $(0.55358, -0.55358)$: global minimum.





    share|cite|improve this answer

























      up vote
      0
      down vote













      Setting the gradient of a multi variable equation equal to zero will give you the (local)minimum, (local)maximum or saddle points.



      In your case, even though the gradient equation system is non-linear, we note that adding the two equations we get $$8x_1 + 8x_2 = 0 Rightarrow x_1=-x_2.$$



      Now we just substitute it in either the first or the second equation of the gradient system. Doing it for the first equation yields
      $$-4(x_2+x_2)^3+8x_2-1=0Rightarrow-32x_2^3+8x_2-1=0$$
      Solving it (using numerical methods, for example) and using the fact that $x_1 = -x_2$, we get the following points:
      $$(-0.13480, 0.13480), quad (-0.41878, 0.41878), quad (0.55358, -0.55358).$$



      This is the contour plot of $F(x)$
      enter image description here



      Analyzing this plot (or doing it numerically) we conclude that:



      • $(-0.13480, 0.13480)$: saddle point;

      • $(-0.41878, 0.41878)$: local minimum;

      • $(0.55358, -0.55358)$: global minimum.





      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Setting the gradient of a multi variable equation equal to zero will give you the (local)minimum, (local)maximum or saddle points.



        In your case, even though the gradient equation system is non-linear, we note that adding the two equations we get $$8x_1 + 8x_2 = 0 Rightarrow x_1=-x_2.$$



        Now we just substitute it in either the first or the second equation of the gradient system. Doing it for the first equation yields
        $$-4(x_2+x_2)^3+8x_2-1=0Rightarrow-32x_2^3+8x_2-1=0$$
        Solving it (using numerical methods, for example) and using the fact that $x_1 = -x_2$, we get the following points:
        $$(-0.13480, 0.13480), quad (-0.41878, 0.41878), quad (0.55358, -0.55358).$$



        This is the contour plot of $F(x)$
        enter image description here



        Analyzing this plot (or doing it numerically) we conclude that:



        • $(-0.13480, 0.13480)$: saddle point;

        • $(-0.41878, 0.41878)$: local minimum;

        • $(0.55358, -0.55358)$: global minimum.





        share|cite|improve this answer













        Setting the gradient of a multi variable equation equal to zero will give you the (local)minimum, (local)maximum or saddle points.



        In your case, even though the gradient equation system is non-linear, we note that adding the two equations we get $$8x_1 + 8x_2 = 0 Rightarrow x_1=-x_2.$$



        Now we just substitute it in either the first or the second equation of the gradient system. Doing it for the first equation yields
        $$-4(x_2+x_2)^3+8x_2-1=0Rightarrow-32x_2^3+8x_2-1=0$$
        Solving it (using numerical methods, for example) and using the fact that $x_1 = -x_2$, we get the following points:
        $$(-0.13480, 0.13480), quad (-0.41878, 0.41878), quad (0.55358, -0.55358).$$



        This is the contour plot of $F(x)$
        enter image description here



        Analyzing this plot (or doing it numerically) we conclude that:



        • $(-0.13480, 0.13480)$: saddle point;

        • $(-0.41878, 0.41878)$: local minimum;

        • $(0.55358, -0.55358)$: global minimum.






        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 22 at 14:54









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