How to study the critical points of a $2$-variable function?
$begingroup$
I am revising some past exam questions and there is one that states:
Study the critical points of the function:
$$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$
According to my professor, this is what I have to do:
Find the gradient, which is $$nabla f(x,y)=begin{bmatrix} 2x-4x^3-4xy^2 \ 2y-4y^3-4x^2y end{bmatrix}.$$
Find the Hessian, using partial derivatives, which is
begin{align*}
f_{xx}(x,y) &= 2-12x^2-4y \
f_{xy}(x,y) &= -8xy \
f_{yx}(x,y) &= -8xy \
f_{yy}(x,y) &= 2-12y^2-4x^2.
end{align*}
And until here it's all good. However, what confuses me is the next step.
We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $nabla f(x,y) = 0$, namely
$$2x(1-2x^2-2y^2)=0 quad text{and} quad 2y(1-2x^2-2y^2)=0.$$
And here I'm lost. On the answer sheet it states:
$$C*f*={(x,y)inmathbb{R}^2 : x^2+y^2=1/2} cup {(0,0)}.$$
What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $cup {(0,0)}$ mean? Is $cup {(0,0)}$ the critical points, which are $0$?
Please explain in the clearest and simpelst way possible!
Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.
Thanks, guys!
multivariable-calculus derivatives maxima-minima hessian-matrix
$endgroup$
add a comment |
$begingroup$
I am revising some past exam questions and there is one that states:
Study the critical points of the function:
$$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$
According to my professor, this is what I have to do:
Find the gradient, which is $$nabla f(x,y)=begin{bmatrix} 2x-4x^3-4xy^2 \ 2y-4y^3-4x^2y end{bmatrix}.$$
Find the Hessian, using partial derivatives, which is
begin{align*}
f_{xx}(x,y) &= 2-12x^2-4y \
f_{xy}(x,y) &= -8xy \
f_{yx}(x,y) &= -8xy \
f_{yy}(x,y) &= 2-12y^2-4x^2.
end{align*}
And until here it's all good. However, what confuses me is the next step.
We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $nabla f(x,y) = 0$, namely
$$2x(1-2x^2-2y^2)=0 quad text{and} quad 2y(1-2x^2-2y^2)=0.$$
And here I'm lost. On the answer sheet it states:
$$C*f*={(x,y)inmathbb{R}^2 : x^2+y^2=1/2} cup {(0,0)}.$$
What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $cup {(0,0)}$ mean? Is $cup {(0,0)}$ the critical points, which are $0$?
Please explain in the clearest and simpelst way possible!
Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.
Thanks, guys!
multivariable-calculus derivatives maxima-minima hessian-matrix
$endgroup$
add a comment |
$begingroup$
I am revising some past exam questions and there is one that states:
Study the critical points of the function:
$$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$
According to my professor, this is what I have to do:
Find the gradient, which is $$nabla f(x,y)=begin{bmatrix} 2x-4x^3-4xy^2 \ 2y-4y^3-4x^2y end{bmatrix}.$$
Find the Hessian, using partial derivatives, which is
begin{align*}
f_{xx}(x,y) &= 2-12x^2-4y \
f_{xy}(x,y) &= -8xy \
f_{yx}(x,y) &= -8xy \
f_{yy}(x,y) &= 2-12y^2-4x^2.
end{align*}
And until here it's all good. However, what confuses me is the next step.
We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $nabla f(x,y) = 0$, namely
$$2x(1-2x^2-2y^2)=0 quad text{and} quad 2y(1-2x^2-2y^2)=0.$$
And here I'm lost. On the answer sheet it states:
$$C*f*={(x,y)inmathbb{R}^2 : x^2+y^2=1/2} cup {(0,0)}.$$
What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $cup {(0,0)}$ mean? Is $cup {(0,0)}$ the critical points, which are $0$?
Please explain in the clearest and simpelst way possible!
Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.
Thanks, guys!
multivariable-calculus derivatives maxima-minima hessian-matrix
$endgroup$
I am revising some past exam questions and there is one that states:
Study the critical points of the function:
$$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$
According to my professor, this is what I have to do:
Find the gradient, which is $$nabla f(x,y)=begin{bmatrix} 2x-4x^3-4xy^2 \ 2y-4y^3-4x^2y end{bmatrix}.$$
Find the Hessian, using partial derivatives, which is
begin{align*}
f_{xx}(x,y) &= 2-12x^2-4y \
f_{xy}(x,y) &= -8xy \
f_{yx}(x,y) &= -8xy \
f_{yy}(x,y) &= 2-12y^2-4x^2.
end{align*}
And until here it's all good. However, what confuses me is the next step.
We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $nabla f(x,y) = 0$, namely
$$2x(1-2x^2-2y^2)=0 quad text{and} quad 2y(1-2x^2-2y^2)=0.$$
And here I'm lost. On the answer sheet it states:
$$C*f*={(x,y)inmathbb{R}^2 : x^2+y^2=1/2} cup {(0,0)}.$$
What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $cup {(0,0)}$ mean? Is $cup {(0,0)}$ the critical points, which are $0$?
Please explain in the clearest and simpelst way possible!
Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.
Thanks, guys!
multivariable-calculus derivatives maxima-minima hessian-matrix
multivariable-calculus derivatives maxima-minima hessian-matrix
edited Dec 8 '18 at 14:05
Rócherz
2,7762721
2,7762721
asked Dec 8 '18 at 13:55
BM97BM97
758
758
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$cup$ means union.
From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
$$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$
It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.
Suppose not, then we must have $x=y=0$.
$endgroup$
$begingroup$
Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
$endgroup$
– BM97
Dec 8 '18 at 14:10
$begingroup$
Yes, they come from those places that you mentioned.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:12
$begingroup$
Thank you !!! You are great!
$endgroup$
– BM97
Dec 8 '18 at 14:13
$begingroup$
May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
$endgroup$
– BM97
Dec 8 '18 at 14:41
$begingroup$
The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:43
|
show 3 more comments
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1 Answer
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1 Answer
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votes
$begingroup$
$cup$ means union.
From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
$$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$
It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.
Suppose not, then we must have $x=y=0$.
$endgroup$
$begingroup$
Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
$endgroup$
– BM97
Dec 8 '18 at 14:10
$begingroup$
Yes, they come from those places that you mentioned.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:12
$begingroup$
Thank you !!! You are great!
$endgroup$
– BM97
Dec 8 '18 at 14:13
$begingroup$
May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
$endgroup$
– BM97
Dec 8 '18 at 14:41
$begingroup$
The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:43
|
show 3 more comments
$begingroup$
$cup$ means union.
From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
$$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$
It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.
Suppose not, then we must have $x=y=0$.
$endgroup$
$begingroup$
Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
$endgroup$
– BM97
Dec 8 '18 at 14:10
$begingroup$
Yes, they come from those places that you mentioned.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:12
$begingroup$
Thank you !!! You are great!
$endgroup$
– BM97
Dec 8 '18 at 14:13
$begingroup$
May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
$endgroup$
– BM97
Dec 8 '18 at 14:41
$begingroup$
The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:43
|
show 3 more comments
$begingroup$
$cup$ means union.
From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
$$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$
It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.
Suppose not, then we must have $x=y=0$.
$endgroup$
$cup$ means union.
From $$2x(1-2x^2-2y^2)=0 iff 2x=0 text{ or } 1-2x^2-2y^2=0$$
$$2y(1-2x^2-2y^2)=0 iff 2y=0 text{ or } 1-2x^2-2y^2=0$$
It is possible that $1-2x^2-2y^2=0 iff x^2+y^2 = frac12$.
Suppose not, then we must have $x=y=0$.
answered Dec 8 '18 at 13:59
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
$begingroup$
Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
$endgroup$
– BM97
Dec 8 '18 at 14:10
$begingroup$
Yes, they come from those places that you mentioned.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:12
$begingroup$
Thank you !!! You are great!
$endgroup$
– BM97
Dec 8 '18 at 14:13
$begingroup$
May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
$endgroup$
– BM97
Dec 8 '18 at 14:41
$begingroup$
The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:43
|
show 3 more comments
$begingroup$
Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
$endgroup$
– BM97
Dec 8 '18 at 14:10
$begingroup$
Yes, they come from those places that you mentioned.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:12
$begingroup$
Thank you !!! You are great!
$endgroup$
– BM97
Dec 8 '18 at 14:13
$begingroup$
May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
$endgroup$
– BM97
Dec 8 '18 at 14:41
$begingroup$
The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:43
$begingroup$
Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
$endgroup$
– BM97
Dec 8 '18 at 14:10
$begingroup$
Thanks for the explanation ! So this means that I the U(0,0) is coming from the 2x=0 and the 2y=0, and that the x^2+y^2=1/2 is coming from the 1-2x^2-2y^2, just by re-arranging the equation ? I'm sorry for the too many questions, I have dyscalculia and I need to get even the simplest things straight!
$endgroup$
– BM97
Dec 8 '18 at 14:10
$begingroup$
Yes, they come from those places that you mentioned.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:12
$begingroup$
Yes, they come from those places that you mentioned.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:12
$begingroup$
Thank you !!! You are great!
$endgroup$
– BM97
Dec 8 '18 at 14:13
$begingroup$
Thank you !!! You are great!
$endgroup$
– BM97
Dec 8 '18 at 14:13
$begingroup$
May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
$endgroup$
– BM97
Dec 8 '18 at 14:41
$begingroup$
May i ask you one last thing? What we wrote before implies that that the set of critical points of f(x, y) is given by the origin and the circle or radius √1/2 centered at the origin. When it says to study the hessian at the origin, the matrix Hf(0,0) is a 2x2 matrix , with on top 2,0 and on botthom 0,2.... may i ask you where the 2's are coming from or is it a general formula?
$endgroup$
– BM97
Dec 8 '18 at 14:41
$begingroup$
The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:43
$begingroup$
The top $2$ at the bottom corresponds to computing $f_{xx}$ and evaluate it at $(0,0)$. The bottom $2$ corresponds to computing $f_{yy}$ and evaluate it at $(0,0)$.
$endgroup$
– Siong Thye Goh
Dec 8 '18 at 14:43
|
show 3 more comments
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