Finding all stationary points quickly












1












$begingroup$


I need to find all the stationary points of the following function defined on $mathbb{R}^2$:
$$f(x_1, x_2) = (x_2 - x_1^2)(x_1 - x_2^2)$$
I would like to know if in general there is a quick way of solving the resulting non linear systems in $mathbb{R}^2$ where powers go up to $3$rd or $4$th order.



How I Solved it



Obviously the first step is to find the gradient and set it to zero to find the system of equations.
$$nabla f(x_1, x_2) = left(x_2-3x_1^2+2x_1x_2^2,, , ,, x_1-3x_2^2+2x_1^2x_2right)^T =0$$
Then since it looks symmetric in $x_1$ and $x_2$ I guessed a solution for $x_1=x_2$, thus:
$$
begin{cases}
x_1-3x_1^2+2x_1x_1^2 = 0 \
x_1-3x_1^2+2x_1^2x_1 = 0
end{cases}
Longrightarrow x_1(2x_1-3x_1 +1)=0 Longrightarrow x_1(2x_1-1)(x_1-1)=0
$$
From which we obtain $A=(0,0)^T$, $B=left(frac{1}{2},frac{1}{2}right)^T$, $C=(1, 1)^T$. Calculating the Hessian we see that two of the are saddle points and one is a maximum. However now I don't know if there are any more saddle points, and in general when using symmetry arguments I don't know how to prove that these are all the possible solutions. I know there is a different method to solve this almost as quickly as this but that guarantees to find all solutions (there has to cause the exercise uses a non-symmetric method)






calculus linear-algebra algebra-precalculus multivariable-calculus optimization







share|cite|improve this question











$endgroup$












  • $begingroup$
    By third or fourth order I mean that there are many similar problems where usually the powers go up to that order
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:08
















1












$begingroup$


I need to find all the stationary points of the following function defined on $mathbb{R}^2$:
$$f(x_1, x_2) = (x_2 - x_1^2)(x_1 - x_2^2)$$
I would like to know if in general there is a quick way of solving the resulting non linear systems in $mathbb{R}^2$ where powers go up to $3$rd or $4$th order.



How I Solved it



Obviously the first step is to find the gradient and set it to zero to find the system of equations.
$$nabla f(x_1, x_2) = left(x_2-3x_1^2+2x_1x_2^2,, , ,, x_1-3x_2^2+2x_1^2x_2right)^T =0$$
Then since it looks symmetric in $x_1$ and $x_2$ I guessed a solution for $x_1=x_2$, thus:
$$
begin{cases}
x_1-3x_1^2+2x_1x_1^2 = 0 \
x_1-3x_1^2+2x_1^2x_1 = 0
end{cases}
Longrightarrow x_1(2x_1-3x_1 +1)=0 Longrightarrow x_1(2x_1-1)(x_1-1)=0
$$
From which we obtain $A=(0,0)^T$, $B=left(frac{1}{2},frac{1}{2}right)^T$, $C=(1, 1)^T$. Calculating the Hessian we see that two of the are saddle points and one is a maximum. However now I don't know if there are any more saddle points, and in general when using symmetry arguments I don't know how to prove that these are all the possible solutions. I know there is a different method to solve this almost as quickly as this but that guarantees to find all solutions (there has to cause the exercise uses a non-symmetric method)






calculus linear-algebra algebra-precalculus multivariable-calculus optimization







share|cite|improve this question











$endgroup$












  • $begingroup$
    By third or fourth order I mean that there are many similar problems where usually the powers go up to that order
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:08














1












1








1


1



$begingroup$


I need to find all the stationary points of the following function defined on $mathbb{R}^2$:
$$f(x_1, x_2) = (x_2 - x_1^2)(x_1 - x_2^2)$$
I would like to know if in general there is a quick way of solving the resulting non linear systems in $mathbb{R}^2$ where powers go up to $3$rd or $4$th order.



How I Solved it



Obviously the first step is to find the gradient and set it to zero to find the system of equations.
$$nabla f(x_1, x_2) = left(x_2-3x_1^2+2x_1x_2^2,, , ,, x_1-3x_2^2+2x_1^2x_2right)^T =0$$
Then since it looks symmetric in $x_1$ and $x_2$ I guessed a solution for $x_1=x_2$, thus:
$$
begin{cases}
x_1-3x_1^2+2x_1x_1^2 = 0 \
x_1-3x_1^2+2x_1^2x_1 = 0
end{cases}
Longrightarrow x_1(2x_1-3x_1 +1)=0 Longrightarrow x_1(2x_1-1)(x_1-1)=0
$$
From which we obtain $A=(0,0)^T$, $B=left(frac{1}{2},frac{1}{2}right)^T$, $C=(1, 1)^T$. Calculating the Hessian we see that two of the are saddle points and one is a maximum. However now I don't know if there are any more saddle points, and in general when using symmetry arguments I don't know how to prove that these are all the possible solutions. I know there is a different method to solve this almost as quickly as this but that guarantees to find all solutions (there has to cause the exercise uses a non-symmetric method)






calculus linear-algebra algebra-precalculus multivariable-calculus optimization







share|cite|improve this question











$endgroup$




I need to find all the stationary points of the following function defined on $mathbb{R}^2$:
$$f(x_1, x_2) = (x_2 - x_1^2)(x_1 - x_2^2)$$
I would like to know if in general there is a quick way of solving the resulting non linear systems in $mathbb{R}^2$ where powers go up to $3$rd or $4$th order.



How I Solved it



Obviously the first step is to find the gradient and set it to zero to find the system of equations.
$$nabla f(x_1, x_2) = left(x_2-3x_1^2+2x_1x_2^2,, , ,, x_1-3x_2^2+2x_1^2x_2right)^T =0$$
Then since it looks symmetric in $x_1$ and $x_2$ I guessed a solution for $x_1=x_2$, thus:
$$
begin{cases}
x_1-3x_1^2+2x_1x_1^2 = 0 \
x_1-3x_1^2+2x_1^2x_1 = 0
end{cases}
Longrightarrow x_1(2x_1-3x_1 +1)=0 Longrightarrow x_1(2x_1-1)(x_1-1)=0
$$
From which we obtain $A=(0,0)^T$, $B=left(frac{1}{2},frac{1}{2}right)^T$, $C=(1, 1)^T$. Calculating the Hessian we see that two of the are saddle points and one is a maximum. However now I don't know if there are any more saddle points, and in general when using symmetry arguments I don't know how to prove that these are all the possible solutions. I know there is a different method to solve this almost as quickly as this but that guarantees to find all solutions (there has to cause the exercise uses a non-symmetric method)






calculus linear-algebra algebra-precalculus multivariable-calculus optimization




calculus linear-algebra algebra-precalculus multivariable-calculus optimization






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edited Dec 29 '18 at 14:14







Euler_Salter

















asked Dec 29 '18 at 13:57









Euler_SalterEuler_Salter

2,0581437




2,0581437












  • $begingroup$
    By third or fourth order I mean that there are many similar problems where usually the powers go up to that order
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:08


















  • $begingroup$
    By third or fourth order I mean that there are many similar problems where usually the powers go up to that order
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:08
















$begingroup$
By third or fourth order I mean that there are many similar problems where usually the powers go up to that order
$endgroup$
– Euler_Salter
Dec 29 '18 at 14:08




$begingroup$
By third or fourth order I mean that there are many similar problems where usually the powers go up to that order
$endgroup$
– Euler_Salter
Dec 29 '18 at 14:08










1 Answer
1






active

oldest

votes


















0












$begingroup$

Actually we get $$P(1;1)$$ and $$P(0;0)$$ are both saddle points and $$P(frac{1}{2};frac{1}{2})$$ is a maximum with $$f(x_1,x_2)geq frac{1}{16}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    yes that's true, has eigenvalues $-4.5$ and $-0.5$, I made a mistake
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:13










  • $begingroup$
    Still, this is not the point of the answer, I tried guessing the definiteness of the Hessian without finding eigenvalues
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:14






  • 2




    $begingroup$
    Ok, nice that i could help you!
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 29 '18 at 14:14












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1 Answer
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1 Answer
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active

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$begingroup$

Actually we get $$P(1;1)$$ and $$P(0;0)$$ are both saddle points and $$P(frac{1}{2};frac{1}{2})$$ is a maximum with $$f(x_1,x_2)geq frac{1}{16}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    yes that's true, has eigenvalues $-4.5$ and $-0.5$, I made a mistake
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:13










  • $begingroup$
    Still, this is not the point of the answer, I tried guessing the definiteness of the Hessian without finding eigenvalues
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:14






  • 2




    $begingroup$
    Ok, nice that i could help you!
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 29 '18 at 14:14
















0












$begingroup$

Actually we get $$P(1;1)$$ and $$P(0;0)$$ are both saddle points and $$P(frac{1}{2};frac{1}{2})$$ is a maximum with $$f(x_1,x_2)geq frac{1}{16}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    yes that's true, has eigenvalues $-4.5$ and $-0.5$, I made a mistake
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:13










  • $begingroup$
    Still, this is not the point of the answer, I tried guessing the definiteness of the Hessian without finding eigenvalues
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:14






  • 2




    $begingroup$
    Ok, nice that i could help you!
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 29 '18 at 14:14














0












0








0





$begingroup$

Actually we get $$P(1;1)$$ and $$P(0;0)$$ are both saddle points and $$P(frac{1}{2};frac{1}{2})$$ is a maximum with $$f(x_1,x_2)geq frac{1}{16}$$






share|cite|improve this answer









$endgroup$



Actually we get $$P(1;1)$$ and $$P(0;0)$$ are both saddle points and $$P(frac{1}{2};frac{1}{2})$$ is a maximum with $$f(x_1,x_2)geq frac{1}{16}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 14:10









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867




79k42867












  • $begingroup$
    yes that's true, has eigenvalues $-4.5$ and $-0.5$, I made a mistake
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:13










  • $begingroup$
    Still, this is not the point of the answer, I tried guessing the definiteness of the Hessian without finding eigenvalues
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:14






  • 2




    $begingroup$
    Ok, nice that i could help you!
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 29 '18 at 14:14


















  • $begingroup$
    yes that's true, has eigenvalues $-4.5$ and $-0.5$, I made a mistake
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:13










  • $begingroup$
    Still, this is not the point of the answer, I tried guessing the definiteness of the Hessian without finding eigenvalues
    $endgroup$
    – Euler_Salter
    Dec 29 '18 at 14:14






  • 2




    $begingroup$
    Ok, nice that i could help you!
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 29 '18 at 14:14
















$begingroup$
yes that's true, has eigenvalues $-4.5$ and $-0.5$, I made a mistake
$endgroup$
– Euler_Salter
Dec 29 '18 at 14:13




$begingroup$
yes that's true, has eigenvalues $-4.5$ and $-0.5$, I made a mistake
$endgroup$
– Euler_Salter
Dec 29 '18 at 14:13












$begingroup$
Still, this is not the point of the answer, I tried guessing the definiteness of the Hessian without finding eigenvalues
$endgroup$
– Euler_Salter
Dec 29 '18 at 14:14




$begingroup$
Still, this is not the point of the answer, I tried guessing the definiteness of the Hessian without finding eigenvalues
$endgroup$
– Euler_Salter
Dec 29 '18 at 14:14




2




2




$begingroup$
Ok, nice that i could help you!
$endgroup$
– Dr. Sonnhard Graubner
Dec 29 '18 at 14:14




$begingroup$
Ok, nice that i could help you!
$endgroup$
– Dr. Sonnhard Graubner
Dec 29 '18 at 14:14


















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