When can I say my 2D mapping is really 1D?
$begingroup$
Consider the following mapping:
$$left(begin{matrix}x\
y
end{matrix}right)mapstoleft(begin{matrix}x+y\
alpha y+sinleft(2pileft(x+yright)right)
end{matrix}right)$$
with $x,yinmathbb{R}$ and the parameter $alphainleft[-1,1right]$
For $alpha=0$ this reduces to
$$left(begin{matrix}x\
y
end{matrix}right)mapstoleft(begin{matrix}x+y\
sinleft(2pileft(x+yright)right)
end{matrix}right)$$
I want to say something along the lines of "This is now effectively a one dimensional map", but in what sense is that true? I mean, formally this is still a mapping from $mathbb{R}^{2}$ to $mathbb{R}^{2}$.
So how can I express this intuition in a more rigorous way?
real-analysis functions
asked Dec 6 '18 at 20:11
H.RappeportH.Rappeport
6761510
$endgroup$
add a comment |
$begingroup$
Consider the following mapping:
$$left(begin{matrix}x\
y
end{matrix}right)mapstoleft(begin{matrix}x+y\
alpha y+sinleft(2pileft(x+yright)right)
end{matrix}right)$$
with $x,yinmathbb{R}$ and the parameter $alphainleft[-1,1right]$
For $alpha=0$ this reduces to
$$left(begin{matrix}x\
y
end{matrix}right)mapstoleft(begin{matrix}x+y\
sinleft(2pileft(x+yright)right)
end{matrix}right)$$
I want to say something along the lines of "This is now effectively a one dimensional map", but in what sense is that true? I mean, formally this is still a mapping from $mathbb{R}^{2}$ to $mathbb{R}^{2}$.
So how can I express this intuition in a more rigorous way?
real-analysis functions
asked Dec 6 '18 at 20:11
H.RappeportH.Rappeport
6761510
$endgroup$
$begingroup$
What are you trying to say with the phrase “a one-dimensional map” here? With $alpha=0$, the image of the plane is indeed one-dimensional, but it’s still a map from $mathbb R^2$ to $mathbb R^2$ as you’ve noted.
$endgroup$
– amd
Dec 6 '18 at 20:14
$begingroup$
The map "factors through" $mathbb R$. There are a few questions on the site about the meaning of "factor through".
$endgroup$
– Mark S.
Dec 6 '18 at 21:08
$begingroup$
maybe consider that the determinant of the Jacobian vanishes everywhere. If it is zero, any 2-d disk is mapped to a zero area patch. Is that "1-d" enough for you?
$endgroup$
– user617446
Dec 31 '18 at 14:17
add a comment |
$begingroup$
Consider the following mapping:
$$left(begin{matrix}x\
y
end{matrix}right)mapstoleft(begin{matrix}x+y\
alpha y+sinleft(2pileft(x+yright)right)
end{matrix}right)$$
with $x,yinmathbb{R}$ and the parameter $alphainleft[-1,1right]$
For $alpha=0$ this reduces to
$$left(begin{matrix}x\
y
end{matrix}right)mapstoleft(begin{matrix}x+y\
sinleft(2pileft(x+yright)right)
end{matrix}right)$$
I want to say something along the lines of "This is now effectively a one dimensional map", but in what sense is that true? I mean, formally this is still a mapping from $mathbb{R}^{2}$ to $mathbb{R}^{2}$.
So how can I express this intuition in a more rigorous way?
real-analysis functions
asked Dec 6 '18 at 20:11
H.RappeportH.Rappeport
6761510
$endgroup$
Consider the following mapping:
$$left(begin{matrix}x\
y
end{matrix}right)mapstoleft(begin{matrix}x+y\
alpha y+sinleft(2pileft(x+yright)right)
end{matrix}right)$$
with $x,yinmathbb{R}$ and the parameter $alphainleft[-1,1right]$
For $alpha=0$ this reduces to
$$left(begin{matrix}x\
y
end{matrix}right)mapstoleft(begin{matrix}x+y\
sinleft(2pileft(x+yright)right)
end{matrix}right)$$
I want to say something along the lines of "This is now effectively a one dimensional map", but in what sense is that true? I mean, formally this is still a mapping from $mathbb{R}^{2}$ to $mathbb{R}^{2}$.
So how can I express this intuition in a more rigorous way?
real-analysis functions
real-analysis functions
asked Dec 6 '18 at 20:11
H.RappeportH.Rappeport
6761510
asked Dec 6 '18 at 20:11
H.RappeportH.Rappeport
6761510
asked Dec 6 '18 at 20:11
H.RappeportH.Rappeport
6761510
asked Dec 6 '18 at 20:11
H.RappeportH.Rappeport
6761510
asked Dec 6 '18 at 20:11
H.RappeportH.Rappeport
6761510
6761510
$begingroup$
What are you trying to say with the phrase “a one-dimensional map” here? With $alpha=0$, the image of the plane is indeed one-dimensional, but it’s still a map from $mathbb R^2$ to $mathbb R^2$ as you’ve noted.
$endgroup$
– amd
Dec 6 '18 at 20:14
$begingroup$
The map "factors through" $mathbb R$. There are a few questions on the site about the meaning of "factor through".
$endgroup$
– Mark S.
Dec 6 '18 at 21:08
$begingroup$
maybe consider that the determinant of the Jacobian vanishes everywhere. If it is zero, any 2-d disk is mapped to a zero area patch. Is that "1-d" enough for you?
$endgroup$
– user617446
Dec 31 '18 at 14:17
add a comment |
$begingroup$
What are you trying to say with the phrase “a one-dimensional map” here? With $alpha=0$, the image of the plane is indeed one-dimensional, but it’s still a map from $mathbb R^2$ to $mathbb R^2$ as you’ve noted.
$endgroup$
– amd
Dec 6 '18 at 20:14
$begingroup$
The map "factors through" $mathbb R$. There are a few questions on the site about the meaning of "factor through".
$endgroup$
– Mark S.
Dec 6 '18 at 21:08
$begingroup$
maybe consider that the determinant of the Jacobian vanishes everywhere. If it is zero, any 2-d disk is mapped to a zero area patch. Is that "1-d" enough for you?
$endgroup$
– user617446
Dec 31 '18 at 14:17
$begingroup$
What are you trying to say with the phrase “a one-dimensional map” here? With $alpha=0$, the image of the plane is indeed one-dimensional, but it’s still a map from $mathbb R^2$ to $mathbb R^2$ as you’ve noted.
$endgroup$
– amd
Dec 6 '18 at 20:14
$begingroup$
What are you trying to say with the phrase “a one-dimensional map” here? With $alpha=0$, the image of the plane is indeed one-dimensional, but it’s still a map from $mathbb R^2$ to $mathbb R^2$ as you’ve noted.
$endgroup$
– amd
Dec 6 '18 at 20:14
$begingroup$
The map "factors through" $mathbb R$. There are a few questions on the site about the meaning of "factor through".
$endgroup$
– Mark S.
Dec 6 '18 at 21:08
$begingroup$
The map "factors through" $mathbb R$. There are a few questions on the site about the meaning of "factor through".
$endgroup$
– Mark S.
Dec 6 '18 at 21:08
$begingroup$
maybe consider that the determinant of the Jacobian vanishes everywhere. If it is zero, any 2-d disk is mapped to a zero area patch. Is that "1-d" enough for you?
$endgroup$
– user617446
Dec 31 '18 at 14:17
$begingroup$
maybe consider that the determinant of the Jacobian vanishes everywhere. If it is zero, any 2-d disk is mapped to a zero area patch. Is that "1-d" enough for you?
$endgroup$
– user617446
Dec 31 '18 at 14:17
add a comment |
1 Answer
1
active
oldest
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$begingroup$
I will name the functions in the OP. For each $alphain [0,1]$, set $f_alpha:mathbb R^2tomathbb R^2$ to be given by $begin{pmatrix}x\yend{pmatrix}mapstobegin{pmatrix}x+y\alpha y+sinleft(2pi(x+y)right)end{pmatrix}$.
Then $f_0$ reduces to $begin{pmatrix}x\yend{pmatrix}mapstobegin{pmatrix}x+y\sinleft(2pi(x+y)right)end{pmatrix}$, which has a special property that can be phrased in a few ways.
In an informal context, I might simply say "$f_0$ is a function of $x+y$ alone". Speaking to a mathematician, I might also say "$f_0$ factors through $mathbb R$"*.
To be precise, note that $f_0$ is the composition $hcirc g$ where $g:mathbb R^2 to mathbb R$ is given by $begin{pmatrix}x\yend{pmatrix}mapsto x+y$ and $h:mathbb Rtomathbb R^2$ is given by $tmapstobegin{pmatrix}t\sin(2pi t)end{pmatrix}$.
*This usage of "factor through" is also attested in this MSE answer.
answered Dec 31 '18 at 13:33
Mark S.Mark S.
11.7k22669
$endgroup$
add a comment |
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$begingroup$
I will name the functions in the OP. For each $alphain [0,1]$, set $f_alpha:mathbb R^2tomathbb R^2$ to be given by $begin{pmatrix}x\yend{pmatrix}mapstobegin{pmatrix}x+y\alpha y+sinleft(2pi(x+y)right)end{pmatrix}$.
Then $f_0$ reduces to $begin{pmatrix}x\yend{pmatrix}mapstobegin{pmatrix}x+y\sinleft(2pi(x+y)right)end{pmatrix}$, which has a special property that can be phrased in a few ways.
In an informal context, I might simply say "$f_0$ is a function of $x+y$ alone". Speaking to a mathematician, I might also say "$f_0$ factors through $mathbb R$"*.
To be precise, note that $f_0$ is the composition $hcirc g$ where $g:mathbb R^2 to mathbb R$ is given by $begin{pmatrix}x\yend{pmatrix}mapsto x+y$ and $h:mathbb Rtomathbb R^2$ is given by $tmapstobegin{pmatrix}t\sin(2pi t)end{pmatrix}$.
*This usage of "factor through" is also attested in this MSE answer.
answered Dec 31 '18 at 13:33
Mark S.Mark S.
11.7k22669
$endgroup$
add a comment |
$begingroup$
I will name the functions in the OP. For each $alphain [0,1]$, set $f_alpha:mathbb R^2tomathbb R^2$ to be given by $begin{pmatrix}x\yend{pmatrix}mapstobegin{pmatrix}x+y\alpha y+sinleft(2pi(x+y)right)end{pmatrix}$.
Then $f_0$ reduces to $begin{pmatrix}x\yend{pmatrix}mapstobegin{pmatrix}x+y\sinleft(2pi(x+y)right)end{pmatrix}$, which has a special property that can be phrased in a few ways.
In an informal context, I might simply say "$f_0$ is a function of $x+y$ alone". Speaking to a mathematician, I might also say "$f_0$ factors through $mathbb R$"*.
To be precise, note that $f_0$ is the composition $hcirc g$ where $g:mathbb R^2 to mathbb R$ is given by $begin{pmatrix}x\yend{pmatrix}mapsto x+y$ and $h:mathbb Rtomathbb R^2$ is given by $tmapstobegin{pmatrix}t\sin(2pi t)end{pmatrix}$.
*This usage of "factor through" is also attested in this MSE answer.
answered Dec 31 '18 at 13:33
Mark S.Mark S.
11.7k22669
$endgroup$
add a comment |
$begingroup$
I will name the functions in the OP. For each $alphain [0,1]$, set $f_alpha:mathbb R^2tomathbb R^2$ to be given by $begin{pmatrix}x\yend{pmatrix}mapstobegin{pmatrix}x+y\alpha y+sinleft(2pi(x+y)right)end{pmatrix}$.
Then $f_0$ reduces to $begin{pmatrix}x\yend{pmatrix}mapstobegin{pmatrix}x+y\sinleft(2pi(x+y)right)end{pmatrix}$, which has a special property that can be phrased in a few ways.
In an informal context, I might simply say "$f_0$ is a function of $x+y$ alone". Speaking to a mathematician, I might also say "$f_0$ factors through $mathbb R$"*.
To be precise, note that $f_0$ is the composition $hcirc g$ where $g:mathbb R^2 to mathbb R$ is given by $begin{pmatrix}x\yend{pmatrix}mapsto x+y$ and $h:mathbb Rtomathbb R^2$ is given by $tmapstobegin{pmatrix}t\sin(2pi t)end{pmatrix}$.
*This usage of "factor through" is also attested in this MSE answer.
answered Dec 31 '18 at 13:33
Mark S.Mark S.
11.7k22669
$endgroup$
I will name the functions in the OP. For each $alphain [0,1]$, set $f_alpha:mathbb R^2tomathbb R^2$ to be given by $begin{pmatrix}x\yend{pmatrix}mapstobegin{pmatrix}x+y\alpha y+sinleft(2pi(x+y)right)end{pmatrix}$.
Then $f_0$ reduces to $begin{pmatrix}x\yend{pmatrix}mapstobegin{pmatrix}x+y\sinleft(2pi(x+y)right)end{pmatrix}$, which has a special property that can be phrased in a few ways.
In an informal context, I might simply say "$f_0$ is a function of $x+y$ alone". Speaking to a mathematician, I might also say "$f_0$ factors through $mathbb R$"*.
To be precise, note that $f_0$ is the composition $hcirc g$ where $g:mathbb R^2 to mathbb R$ is given by $begin{pmatrix}x\yend{pmatrix}mapsto x+y$ and $h:mathbb Rtomathbb R^2$ is given by $tmapstobegin{pmatrix}t\sin(2pi t)end{pmatrix}$.
*This usage of "factor through" is also attested in this MSE answer.
answered Dec 31 '18 at 13:33
Mark S.Mark S.
11.7k22669
answered Dec 31 '18 at 13:33
Mark S.Mark S.
11.7k22669
answered Dec 31 '18 at 13:33
Mark S.Mark S.
11.7k22669
answered Dec 31 '18 at 13:33
Mark S.Mark S.
11.7k22669
11.7k22669
add a comment |
add a comment |
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$begingroup$
What are you trying to say with the phrase “a one-dimensional map” here? With $alpha=0$, the image of the plane is indeed one-dimensional, but it’s still a map from $mathbb R^2$ to $mathbb R^2$ as you’ve noted.
$endgroup$
– amd
Dec 6 '18 at 20:14
$begingroup$
The map "factors through" $mathbb R$. There are a few questions on the site about the meaning of "factor through".
$endgroup$
– Mark S.
Dec 6 '18 at 21:08
$begingroup$
maybe consider that the determinant of the Jacobian vanishes everywhere. If it is zero, any 2-d disk is mapped to a zero area patch. Is that "1-d" enough for you?
$endgroup$
– user617446
Dec 31 '18 at 14:17