How do I sketch the level curves of $f(x,y) = x^2 - y^2$.
$begingroup$
I get that the critical point of the function $f(x,y)$ is $(0,0)$ which is a saddle point. I know how the level curves are supposed to look in general for saddle points, but I wanted to know if it's possible to determine which areas surrounding the saddle point would be highest or lowest without resorting to computational methods?
calculus multivariable-calculus
asked Dec 4 '18 at 2:03
K.MK.M
700412
$endgroup$
add a comment |
$begingroup$
I get that the critical point of the function $f(x,y)$ is $(0,0)$ which is a saddle point. I know how the level curves are supposed to look in general for saddle points, but I wanted to know if it's possible to determine which areas surrounding the saddle point would be highest or lowest without resorting to computational methods?
calculus multivariable-calculus
asked Dec 4 '18 at 2:03
K.MK.M
700412
$endgroup$
$begingroup$
"which areas surrounding the saddle point would be the highest or lowest " what does "highest" or "lowest" mean? I mean I have drawn the level curves by hand but there are no local maximas or minimas.
$endgroup$
– Sameer Baheti
Dec 4 '18 at 9:12
add a comment |
$begingroup$
I get that the critical point of the function $f(x,y)$ is $(0,0)$ which is a saddle point. I know how the level curves are supposed to look in general for saddle points, but I wanted to know if it's possible to determine which areas surrounding the saddle point would be highest or lowest without resorting to computational methods?
calculus multivariable-calculus
asked Dec 4 '18 at 2:03
K.MK.M
700412
$endgroup$
I get that the critical point of the function $f(x,y)$ is $(0,0)$ which is a saddle point. I know how the level curves are supposed to look in general for saddle points, but I wanted to know if it's possible to determine which areas surrounding the saddle point would be highest or lowest without resorting to computational methods?
calculus multivariable-calculus
calculus multivariable-calculus
asked Dec 4 '18 at 2:03
K.MK.M
700412
asked Dec 4 '18 at 2:03
K.MK.M
700412
edited Dec 4 '18 at 2:31
K.M
asked Dec 4 '18 at 2:03
K.MK.M
700412
asked Dec 4 '18 at 2:03
K.MK.M
700412
asked Dec 4 '18 at 2:03
K.MK.M
700412
700412
$begingroup$
"which areas surrounding the saddle point would be the highest or lowest " what does "highest" or "lowest" mean? I mean I have drawn the level curves by hand but there are no local maximas or minimas.
$endgroup$
– Sameer Baheti
Dec 4 '18 at 9:12
add a comment |
$begingroup$
"which areas surrounding the saddle point would be the highest or lowest " what does "highest" or "lowest" mean? I mean I have drawn the level curves by hand but there are no local maximas or minimas.
$endgroup$
– Sameer Baheti
Dec 4 '18 at 9:12
$begingroup$
"which areas surrounding the saddle point would be the highest or lowest " what does "highest" or "lowest" mean? I mean I have drawn the level curves by hand but there are no local maximas or minimas.
$endgroup$
– Sameer Baheti
Dec 4 '18 at 9:12
$begingroup$
"which areas surrounding the saddle point would be the highest or lowest " what does "highest" or "lowest" mean? I mean I have drawn the level curves by hand but there are no local maximas or minimas.
$endgroup$
– Sameer Baheti
Dec 4 '18 at 9:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you specifically want to know what the level curves look like in general, you can do no better than solving for $f(x,y)=f(x_*,y_*)+C$ for different constants $C$.
If you just want particular directions which are maximally upwards or downwards, something similar to the second derivative test works.
Consider the matrix of second partial derivatives such that the $i$th row $j$th column entry is the derivative wrt $j$th dimension then the $i$th dimension
$$mathbf{H}f=begin{bmatrix}
frac{partial^2 f}{partial x^2} & frac{partial^2 f}{partial xpartial y}\
frac{partial^2 f}{partial ypartial x} & frac{partial^2 f}{partial y^2}
end{bmatrix}$$
known as the Hessian.
The same way we can write a second order Taylor approximation as
$$f(x)approx f(x_0)+f'(x_0)(x-x_0)+frac{1}{2}f''(x_0)(x-x_0)^2$$
we can write one for a multidimensional function as
$$f(vec{x})approx f(vec{x_0})+nabla f(vec{x_0})cdot(vec{x}-vec{x_0})+frac{1}{2}(vec{x}-vec{x_0})^Tleft(mathbf{H}f(vec{x_0})right)(vec{x}-vec{x_0})$$
where $nabla f$ is the gradient, the vector of first partial derivatives.
When $vec{x_0}$ is a critical point, the gradient vanishes, so locally this looks like a multidimensional quadratic form. To figure out which directions are maximal or minimal, find the eigenvalues and eigenvectors of the Hessian.
Along the direction of an unit eigenvector $vec{v}$ of eigenvalue $lambda$, the function looks like
$$f(x_0+rvec{v})approx f(x_0)+frac{1}{2}rvec{v}^Tleft(mathbf{H}f(vec{x_0})right)rvec{v}=f(x_0)+frac{lambda}{2}r^2$$
which is a parabola, upwards if $lambda>0$.
For your case, since
$$mathbf{H}f(0,0)=begin{bmatrix}2&0\0&-2end{bmatrix}$$
this immediately says that $f(x,y)$ locally curves maximally upwards in the $x$ direction, and maximally downwards in the $y$ direction.
answered Dec 4 '18 at 2:41
obscuransobscurans
1,152311
$endgroup$
$begingroup$
$(x*,y*)$ is the critical point of $f$?
$endgroup$
– K.M
Dec 4 '18 at 3:02
1
$begingroup$
Yeah, the critical point you're trying to draw level curves around
$endgroup$
– obscurans
Dec 4 '18 at 3:14
$begingroup$
In the third paragraph, do you mean $i×i$ dimensional Hessian matrix?
$endgroup$
– K.M
Dec 4 '18 at 3:32
1
$begingroup$
For a function $f:mathbb{R}^nrightarrowmathbb{R}$, the Hessian is a $ntimes n$ matrix. The (i,j) entry of the matrix is the second derivative of $f$ wrt the $i$th input and the $j$th input.
$endgroup$
– obscurans
Dec 4 '18 at 3:35
add a comment |
Your Answer
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$begingroup$
If you specifically want to know what the level curves look like in general, you can do no better than solving for $f(x,y)=f(x_*,y_*)+C$ for different constants $C$.
If you just want particular directions which are maximally upwards or downwards, something similar to the second derivative test works.
Consider the matrix of second partial derivatives such that the $i$th row $j$th column entry is the derivative wrt $j$th dimension then the $i$th dimension
$$mathbf{H}f=begin{bmatrix}
frac{partial^2 f}{partial x^2} & frac{partial^2 f}{partial xpartial y}\
frac{partial^2 f}{partial ypartial x} & frac{partial^2 f}{partial y^2}
end{bmatrix}$$
known as the Hessian.
The same way we can write a second order Taylor approximation as
$$f(x)approx f(x_0)+f'(x_0)(x-x_0)+frac{1}{2}f''(x_0)(x-x_0)^2$$
we can write one for a multidimensional function as
$$f(vec{x})approx f(vec{x_0})+nabla f(vec{x_0})cdot(vec{x}-vec{x_0})+frac{1}{2}(vec{x}-vec{x_0})^Tleft(mathbf{H}f(vec{x_0})right)(vec{x}-vec{x_0})$$
where $nabla f$ is the gradient, the vector of first partial derivatives.
When $vec{x_0}$ is a critical point, the gradient vanishes, so locally this looks like a multidimensional quadratic form. To figure out which directions are maximal or minimal, find the eigenvalues and eigenvectors of the Hessian.
Along the direction of an unit eigenvector $vec{v}$ of eigenvalue $lambda$, the function looks like
$$f(x_0+rvec{v})approx f(x_0)+frac{1}{2}rvec{v}^Tleft(mathbf{H}f(vec{x_0})right)rvec{v}=f(x_0)+frac{lambda}{2}r^2$$
which is a parabola, upwards if $lambda>0$.
For your case, since
$$mathbf{H}f(0,0)=begin{bmatrix}2&0\0&-2end{bmatrix}$$
this immediately says that $f(x,y)$ locally curves maximally upwards in the $x$ direction, and maximally downwards in the $y$ direction.
answered Dec 4 '18 at 2:41
obscuransobscurans
1,152311
$endgroup$
$begingroup$
$(x*,y*)$ is the critical point of $f$?
$endgroup$
– K.M
Dec 4 '18 at 3:02
1
$begingroup$
Yeah, the critical point you're trying to draw level curves around
$endgroup$
– obscurans
Dec 4 '18 at 3:14
$begingroup$
In the third paragraph, do you mean $i×i$ dimensional Hessian matrix?
$endgroup$
– K.M
Dec 4 '18 at 3:32
1
$begingroup$
For a function $f:mathbb{R}^nrightarrowmathbb{R}$, the Hessian is a $ntimes n$ matrix. The (i,j) entry of the matrix is the second derivative of $f$ wrt the $i$th input and the $j$th input.
$endgroup$
– obscurans
Dec 4 '18 at 3:35
add a comment |
$begingroup$
If you specifically want to know what the level curves look like in general, you can do no better than solving for $f(x,y)=f(x_*,y_*)+C$ for different constants $C$.
If you just want particular directions which are maximally upwards or downwards, something similar to the second derivative test works.
Consider the matrix of second partial derivatives such that the $i$th row $j$th column entry is the derivative wrt $j$th dimension then the $i$th dimension
$$mathbf{H}f=begin{bmatrix}
frac{partial^2 f}{partial x^2} & frac{partial^2 f}{partial xpartial y}\
frac{partial^2 f}{partial ypartial x} & frac{partial^2 f}{partial y^2}
end{bmatrix}$$
known as the Hessian.
The same way we can write a second order Taylor approximation as
$$f(x)approx f(x_0)+f'(x_0)(x-x_0)+frac{1}{2}f''(x_0)(x-x_0)^2$$
we can write one for a multidimensional function as
$$f(vec{x})approx f(vec{x_0})+nabla f(vec{x_0})cdot(vec{x}-vec{x_0})+frac{1}{2}(vec{x}-vec{x_0})^Tleft(mathbf{H}f(vec{x_0})right)(vec{x}-vec{x_0})$$
where $nabla f$ is the gradient, the vector of first partial derivatives.
When $vec{x_0}$ is a critical point, the gradient vanishes, so locally this looks like a multidimensional quadratic form. To figure out which directions are maximal or minimal, find the eigenvalues and eigenvectors of the Hessian.
Along the direction of an unit eigenvector $vec{v}$ of eigenvalue $lambda$, the function looks like
$$f(x_0+rvec{v})approx f(x_0)+frac{1}{2}rvec{v}^Tleft(mathbf{H}f(vec{x_0})right)rvec{v}=f(x_0)+frac{lambda}{2}r^2$$
which is a parabola, upwards if $lambda>0$.
For your case, since
$$mathbf{H}f(0,0)=begin{bmatrix}2&0\0&-2end{bmatrix}$$
this immediately says that $f(x,y)$ locally curves maximally upwards in the $x$ direction, and maximally downwards in the $y$ direction.
answered Dec 4 '18 at 2:41
obscuransobscurans
1,152311
$endgroup$
$begingroup$
$(x*,y*)$ is the critical point of $f$?
$endgroup$
– K.M
Dec 4 '18 at 3:02
1
$begingroup$
Yeah, the critical point you're trying to draw level curves around
$endgroup$
– obscurans
Dec 4 '18 at 3:14
$begingroup$
In the third paragraph, do you mean $i×i$ dimensional Hessian matrix?
$endgroup$
– K.M
Dec 4 '18 at 3:32
1
$begingroup$
For a function $f:mathbb{R}^nrightarrowmathbb{R}$, the Hessian is a $ntimes n$ matrix. The (i,j) entry of the matrix is the second derivative of $f$ wrt the $i$th input and the $j$th input.
$endgroup$
– obscurans
Dec 4 '18 at 3:35
add a comment |
$begingroup$
If you specifically want to know what the level curves look like in general, you can do no better than solving for $f(x,y)=f(x_*,y_*)+C$ for different constants $C$.
If you just want particular directions which are maximally upwards or downwards, something similar to the second derivative test works.
Consider the matrix of second partial derivatives such that the $i$th row $j$th column entry is the derivative wrt $j$th dimension then the $i$th dimension
$$mathbf{H}f=begin{bmatrix}
frac{partial^2 f}{partial x^2} & frac{partial^2 f}{partial xpartial y}\
frac{partial^2 f}{partial ypartial x} & frac{partial^2 f}{partial y^2}
end{bmatrix}$$
known as the Hessian.
The same way we can write a second order Taylor approximation as
$$f(x)approx f(x_0)+f'(x_0)(x-x_0)+frac{1}{2}f''(x_0)(x-x_0)^2$$
we can write one for a multidimensional function as
$$f(vec{x})approx f(vec{x_0})+nabla f(vec{x_0})cdot(vec{x}-vec{x_0})+frac{1}{2}(vec{x}-vec{x_0})^Tleft(mathbf{H}f(vec{x_0})right)(vec{x}-vec{x_0})$$
where $nabla f$ is the gradient, the vector of first partial derivatives.
When $vec{x_0}$ is a critical point, the gradient vanishes, so locally this looks like a multidimensional quadratic form. To figure out which directions are maximal or minimal, find the eigenvalues and eigenvectors of the Hessian.
Along the direction of an unit eigenvector $vec{v}$ of eigenvalue $lambda$, the function looks like
$$f(x_0+rvec{v})approx f(x_0)+frac{1}{2}rvec{v}^Tleft(mathbf{H}f(vec{x_0})right)rvec{v}=f(x_0)+frac{lambda}{2}r^2$$
which is a parabola, upwards if $lambda>0$.
For your case, since
$$mathbf{H}f(0,0)=begin{bmatrix}2&0\0&-2end{bmatrix}$$
this immediately says that $f(x,y)$ locally curves maximally upwards in the $x$ direction, and maximally downwards in the $y$ direction.
answered Dec 4 '18 at 2:41
obscuransobscurans
1,152311
$endgroup$
If you specifically want to know what the level curves look like in general, you can do no better than solving for $f(x,y)=f(x_*,y_*)+C$ for different constants $C$.
If you just want particular directions which are maximally upwards or downwards, something similar to the second derivative test works.
Consider the matrix of second partial derivatives such that the $i$th row $j$th column entry is the derivative wrt $j$th dimension then the $i$th dimension
$$mathbf{H}f=begin{bmatrix}
frac{partial^2 f}{partial x^2} & frac{partial^2 f}{partial xpartial y}\
frac{partial^2 f}{partial ypartial x} & frac{partial^2 f}{partial y^2}
end{bmatrix}$$
known as the Hessian.
The same way we can write a second order Taylor approximation as
$$f(x)approx f(x_0)+f'(x_0)(x-x_0)+frac{1}{2}f''(x_0)(x-x_0)^2$$
we can write one for a multidimensional function as
$$f(vec{x})approx f(vec{x_0})+nabla f(vec{x_0})cdot(vec{x}-vec{x_0})+frac{1}{2}(vec{x}-vec{x_0})^Tleft(mathbf{H}f(vec{x_0})right)(vec{x}-vec{x_0})$$
where $nabla f$ is the gradient, the vector of first partial derivatives.
When $vec{x_0}$ is a critical point, the gradient vanishes, so locally this looks like a multidimensional quadratic form. To figure out which directions are maximal or minimal, find the eigenvalues and eigenvectors of the Hessian.
Along the direction of an unit eigenvector $vec{v}$ of eigenvalue $lambda$, the function looks like
$$f(x_0+rvec{v})approx f(x_0)+frac{1}{2}rvec{v}^Tleft(mathbf{H}f(vec{x_0})right)rvec{v}=f(x_0)+frac{lambda}{2}r^2$$
which is a parabola, upwards if $lambda>0$.
For your case, since
$$mathbf{H}f(0,0)=begin{bmatrix}2&0\0&-2end{bmatrix}$$
this immediately says that $f(x,y)$ locally curves maximally upwards in the $x$ direction, and maximally downwards in the $y$ direction.
answered Dec 4 '18 at 2:41
obscuransobscurans
1,152311
edited Dec 4 '18 at 2:47
answered Dec 4 '18 at 2:41
obscuransobscurans
1,152311
answered Dec 4 '18 at 2:41
obscuransobscurans
1,152311
answered Dec 4 '18 at 2:41
obscuransobscurans
1,152311
1,152311
$begingroup$
$(x*,y*)$ is the critical point of $f$?
$endgroup$
– K.M
Dec 4 '18 at 3:02
1
$begingroup$
Yeah, the critical point you're trying to draw level curves around
$endgroup$
– obscurans
Dec 4 '18 at 3:14
$begingroup$
In the third paragraph, do you mean $i×i$ dimensional Hessian matrix?
$endgroup$
– K.M
Dec 4 '18 at 3:32
1
$begingroup$
For a function $f:mathbb{R}^nrightarrowmathbb{R}$, the Hessian is a $ntimes n$ matrix. The (i,j) entry of the matrix is the second derivative of $f$ wrt the $i$th input and the $j$th input.
$endgroup$
– obscurans
Dec 4 '18 at 3:35
add a comment |
$begingroup$
$(x*,y*)$ is the critical point of $f$?
$endgroup$
– K.M
Dec 4 '18 at 3:02
1
$begingroup$
Yeah, the critical point you're trying to draw level curves around
$endgroup$
– obscurans
Dec 4 '18 at 3:14
$begingroup$
In the third paragraph, do you mean $i×i$ dimensional Hessian matrix?
$endgroup$
– K.M
Dec 4 '18 at 3:32
1
$begingroup$
For a function $f:mathbb{R}^nrightarrowmathbb{R}$, the Hessian is a $ntimes n$ matrix. The (i,j) entry of the matrix is the second derivative of $f$ wrt the $i$th input and the $j$th input.
$endgroup$
– obscurans
Dec 4 '18 at 3:35
$begingroup$
$(x*,y*)$ is the critical point of $f$?
$endgroup$
– K.M
Dec 4 '18 at 3:02
$begingroup$
$(x*,y*)$ is the critical point of $f$?
$endgroup$
– K.M
Dec 4 '18 at 3:02
1
1
$begingroup$
Yeah, the critical point you're trying to draw level curves around
$endgroup$
– obscurans
Dec 4 '18 at 3:14
$begingroup$
Yeah, the critical point you're trying to draw level curves around
$endgroup$
– obscurans
Dec 4 '18 at 3:14
$begingroup$
In the third paragraph, do you mean $i×i$ dimensional Hessian matrix?
$endgroup$
– K.M
Dec 4 '18 at 3:32
$begingroup$
In the third paragraph, do you mean $i×i$ dimensional Hessian matrix?
$endgroup$
– K.M
Dec 4 '18 at 3:32
1
1
$begingroup$
For a function $f:mathbb{R}^nrightarrowmathbb{R}$, the Hessian is a $ntimes n$ matrix. The (i,j) entry of the matrix is the second derivative of $f$ wrt the $i$th input and the $j$th input.
$endgroup$
– obscurans
Dec 4 '18 at 3:35
$begingroup$
For a function $f:mathbb{R}^nrightarrowmathbb{R}$, the Hessian is a $ntimes n$ matrix. The (i,j) entry of the matrix is the second derivative of $f$ wrt the $i$th input and the $j$th input.
$endgroup$
– obscurans
Dec 4 '18 at 3:35
add a comment |
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$begingroup$
"which areas surrounding the saddle point would be the highest or lowest " what does "highest" or "lowest" mean? I mean I have drawn the level curves by hand but there are no local maximas or minimas.
$endgroup$
– Sameer Baheti
Dec 4 '18 at 9:12