Can 3D co-ordinates be transferred into 2D co-ordinates?Calibration of an eye tracking device: transformation from known gaze pointsCan you divide a square into 5 equal area regionsHow many different parallelograms can be drawn if given three co-ordinates in 3D Cartesian vector?Proving co-ordinates of an equilateral triangle are integers in a planeTransforming an Ellipse into a HyperbolaProjective of a quadrilateral into a squareHow can I find the largest axis-parallel square that can be inscribed into a linearly transformed unit square?finding the similarity transform connecting two sets of matricesWhat area of math develops an overaching theory of transforms?Area of square with integer co-ordinates
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Can 3D co-ordinates be transferred into 2D co-ordinates?
Calibration of an eye tracking device: transformation from known gaze pointsCan you divide a square into 5 equal area regionsHow many different parallelograms can be drawn if given three co-ordinates in 3D Cartesian vector?Proving co-ordinates of an equilateral triangle are integers in a planeTransforming an Ellipse into a HyperbolaProjective of a quadrilateral into a squareHow can I find the largest axis-parallel square that can be inscribed into a linearly transformed unit square?finding the similarity transform connecting two sets of matricesWhat area of math develops an overaching theory of transforms?Area of square with integer co-ordinates
$begingroup$
Is it possible to transform co-ordinates $(a,b,c)$ into $(x,y) $ such that $(x,y)$ is unique for each $(a,b,c)$ ? $a, b, c, x, y$ are in $BbbR$ .
geometry linear-transformations laplace-transform transformation plane-geometry
edited 2 days ago
dmtri
1,6082521
asked 2 days ago
BJKShahBJKShah
84
$endgroup$
add a comment |
$begingroup$
Is it possible to transform co-ordinates $(a,b,c)$ into $(x,y) $ such that $(x,y)$ is unique for each $(a,b,c)$ ? $a, b, c, x, y$ are in $BbbR$ .
geometry linear-transformations laplace-transform transformation plane-geometry
edited 2 days ago
dmtri
1,6082521
asked 2 days ago
BJKShahBJKShah
84
$endgroup$
$begingroup$
Just to clarify, do you want the map from $Bbb R^3$ to $Bbb R^2$ to be invertible, or just exist? As BJKShah, the former is possible but not if it's linear, whereas the latter is possible even then (such a map is an example of a projection).
$endgroup$
– J.G.
2 days ago
$begingroup$
I want the conversion to be possible in both directions.( How to convert former as you said , any links?) Thanks
$endgroup$
– BJKShah
2 days ago
add a comment |
$begingroup$
Is it possible to transform co-ordinates $(a,b,c)$ into $(x,y) $ such that $(x,y)$ is unique for each $(a,b,c)$ ? $a, b, c, x, y$ are in $BbbR$ .
geometry linear-transformations laplace-transform transformation plane-geometry
edited 2 days ago
dmtri
1,6082521
asked 2 days ago
BJKShahBJKShah
84
$endgroup$
Is it possible to transform co-ordinates $(a,b,c)$ into $(x,y) $ such that $(x,y)$ is unique for each $(a,b,c)$ ? $a, b, c, x, y$ are in $BbbR$ .
geometry linear-transformations laplace-transform transformation plane-geometry
geometry linear-transformations laplace-transform transformation plane-geometry
edited 2 days ago
dmtri
1,6082521
asked 2 days ago
BJKShahBJKShah
84
edited 2 days ago
dmtri
1,6082521
asked 2 days ago
BJKShahBJKShah
84
edited 2 days ago
dmtri
1,6082521
edited 2 days ago
dmtri
1,6082521
edited 2 days ago
dmtri
1,6082521
1,6082521
asked 2 days ago
BJKShahBJKShah
84
asked 2 days ago
BJKShahBJKShah
84
asked 2 days ago
BJKShahBJKShah
84
84
$begingroup$
Just to clarify, do you want the map from $Bbb R^3$ to $Bbb R^2$ to be invertible, or just exist? As BJKShah, the former is possible but not if it's linear, whereas the latter is possible even then (such a map is an example of a projection).
$endgroup$
– J.G.
2 days ago
$begingroup$
I want the conversion to be possible in both directions.( How to convert former as you said , any links?) Thanks
$endgroup$
– BJKShah
2 days ago
add a comment |
$begingroup$
Just to clarify, do you want the map from $Bbb R^3$ to $Bbb R^2$ to be invertible, or just exist? As BJKShah, the former is possible but not if it's linear, whereas the latter is possible even then (such a map is an example of a projection).
$endgroup$
– J.G.
2 days ago
$begingroup$
I want the conversion to be possible in both directions.( How to convert former as you said , any links?) Thanks
$endgroup$
– BJKShah
2 days ago
$begingroup$
Just to clarify, do you want the map from $Bbb R^3$ to $Bbb R^2$ to be invertible, or just exist? As BJKShah, the former is possible but not if it's linear, whereas the latter is possible even then (such a map is an example of a projection).
$endgroup$
– J.G.
2 days ago
$begingroup$
Just to clarify, do you want the map from $Bbb R^3$ to $Bbb R^2$ to be invertible, or just exist? As BJKShah, the former is possible but not if it's linear, whereas the latter is possible even then (such a map is an example of a projection).
$endgroup$
– J.G.
2 days ago
$begingroup$
I want the conversion to be possible in both directions.( How to convert former as you said , any links?) Thanks
$endgroup$
– BJKShah
2 days ago
$begingroup$
I want the conversion to be possible in both directions.( How to convert former as you said , any links?) Thanks
$endgroup$
– BJKShah
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is possible, but it's not pretty, and as VENKITESH says, it's certainly not linear. Here is one way: define $$f(a,b,c)=(x,y)$$ where $x=a$, and $y$ is the alternating decimal expansion of $b$ and $c$. What that means is this:
Take the non-terminating decimal expansions of $b$ and $c$; if necessary, insert leading zeroes so that the integral parts have the same length. For instance, if $b=1234.345$ and $c=76.987698798ldots$, we convert $b$ to its non-terminating form $b=colorred1234.344999999ldots$, and we write $c$ as $colorblue0076.987698798ldots$ to make the lengths before the decimal point equal.
Now we can write $y$ by taking digits from $b$ and $c$ in turn:
$$y=colorred1colorblue0colorred2colorblue0colorred3colorblue7colorred4colorblue6.colorred3colorblue9colorred4colorblue8colorred4colorblue7colorred9colorblue6colorred9colorblue9colorred9colorblue8colorred9colorblue7colorred9colorblue9colorred9colorblue8ldots$$
This is just one way. By the way, the name alternating decimal expansion is something I just made up, so don't take it too seriously.
Edited to add: I see you changed your criteria in a comment. This mapping is an injection, but not a surjection, so it doesn't have an inverse. You can make it surjective, but it's messy, and I'm not going to do it here.
Edited to add: As the OP points out in a comment, I have not define the mapping for $b,cle 0$. My bad! But you can fix this by using the decimal expansions of $e^b$ and $e^c$, which are always strictly greater than zero.
answered 2 days ago
TonyKTonyK
43k356135
$endgroup$
$begingroup$
(1,2,1)=(1,21) but (1,-2,-1) and (1,-2,1)=(?) Thanks for the example.
$endgroup$
– BJKShah
2 days ago
add a comment |
$begingroup$
It's is possible to have a bijection between $mathbbR^3$ and $mathbbR^2$, since it can be shown that $mathbbR$ and $mathbbR^2$, and hence every $mathbbR^n$ has the same cardinality. But a linear or affine isomorphism is not possible if the dimensions are different.
edited 2 days ago
tatan
5,79762760
answered 2 days ago
VENKITESHVENKITESH
36817
$endgroup$
$begingroup$
Can you please explain the last sentence a bit more?
$endgroup$
– BJKShah
2 days ago
$begingroup$
Right. A map f:R^nto R^m will be linear if f(u+v)=f(u)+f(v) and f(au)=af(u) for all scalars a and vectors u,v. We can show that if such an f is a bijection, then m=n. This follows from what is called the Rank Nullity Theorem. An affine map is of the form f+c, for a linear map f and a constant vector c. Again if an affine map is bijective, then we can show that m=n.
$endgroup$
– VENKITESH
2 days ago
$begingroup$
Any good books for studying this?
$endgroup$
– BJKShah
2 days ago
$begingroup$
Calculus and Analytic Geometry by Thomas and Finney, for coordinate geometry, and Linear Algebra by Gilbert Strang for linear algebra.
$endgroup$
– VENKITESH
2 days ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is possible, but it's not pretty, and as VENKITESH says, it's certainly not linear. Here is one way: define $$f(a,b,c)=(x,y)$$ where $x=a$, and $y$ is the alternating decimal expansion of $b$ and $c$. What that means is this:
Take the non-terminating decimal expansions of $b$ and $c$; if necessary, insert leading zeroes so that the integral parts have the same length. For instance, if $b=1234.345$ and $c=76.987698798ldots$, we convert $b$ to its non-terminating form $b=colorred1234.344999999ldots$, and we write $c$ as $colorblue0076.987698798ldots$ to make the lengths before the decimal point equal.
Now we can write $y$ by taking digits from $b$ and $c$ in turn:
$$y=colorred1colorblue0colorred2colorblue0colorred3colorblue7colorred4colorblue6.colorred3colorblue9colorred4colorblue8colorred4colorblue7colorred9colorblue6colorred9colorblue9colorred9colorblue8colorred9colorblue7colorred9colorblue9colorred9colorblue8ldots$$
This is just one way. By the way, the name alternating decimal expansion is something I just made up, so don't take it too seriously.
Edited to add: I see you changed your criteria in a comment. This mapping is an injection, but not a surjection, so it doesn't have an inverse. You can make it surjective, but it's messy, and I'm not going to do it here.
Edited to add: As the OP points out in a comment, I have not define the mapping for $b,cle 0$. My bad! But you can fix this by using the decimal expansions of $e^b$ and $e^c$, which are always strictly greater than zero.
answered 2 days ago
TonyKTonyK
43k356135
$endgroup$
$begingroup$
(1,2,1)=(1,21) but (1,-2,-1) and (1,-2,1)=(?) Thanks for the example.
$endgroup$
– BJKShah
2 days ago
add a comment |
$begingroup$
It is possible, but it's not pretty, and as VENKITESH says, it's certainly not linear. Here is one way: define $$f(a,b,c)=(x,y)$$ where $x=a$, and $y$ is the alternating decimal expansion of $b$ and $c$. What that means is this:
Take the non-terminating decimal expansions of $b$ and $c$; if necessary, insert leading zeroes so that the integral parts have the same length. For instance, if $b=1234.345$ and $c=76.987698798ldots$, we convert $b$ to its non-terminating form $b=colorred1234.344999999ldots$, and we write $c$ as $colorblue0076.987698798ldots$ to make the lengths before the decimal point equal.
Now we can write $y$ by taking digits from $b$ and $c$ in turn:
$$y=colorred1colorblue0colorred2colorblue0colorred3colorblue7colorred4colorblue6.colorred3colorblue9colorred4colorblue8colorred4colorblue7colorred9colorblue6colorred9colorblue9colorred9colorblue8colorred9colorblue7colorred9colorblue9colorred9colorblue8ldots$$
This is just one way. By the way, the name alternating decimal expansion is something I just made up, so don't take it too seriously.
Edited to add: I see you changed your criteria in a comment. This mapping is an injection, but not a surjection, so it doesn't have an inverse. You can make it surjective, but it's messy, and I'm not going to do it here.
Edited to add: As the OP points out in a comment, I have not define the mapping for $b,cle 0$. My bad! But you can fix this by using the decimal expansions of $e^b$ and $e^c$, which are always strictly greater than zero.
answered 2 days ago
TonyKTonyK
43k356135
$endgroup$
$begingroup$
(1,2,1)=(1,21) but (1,-2,-1) and (1,-2,1)=(?) Thanks for the example.
$endgroup$
– BJKShah
2 days ago
add a comment |
$begingroup$
It is possible, but it's not pretty, and as VENKITESH says, it's certainly not linear. Here is one way: define $$f(a,b,c)=(x,y)$$ where $x=a$, and $y$ is the alternating decimal expansion of $b$ and $c$. What that means is this:
Take the non-terminating decimal expansions of $b$ and $c$; if necessary, insert leading zeroes so that the integral parts have the same length. For instance, if $b=1234.345$ and $c=76.987698798ldots$, we convert $b$ to its non-terminating form $b=colorred1234.344999999ldots$, and we write $c$ as $colorblue0076.987698798ldots$ to make the lengths before the decimal point equal.
Now we can write $y$ by taking digits from $b$ and $c$ in turn:
$$y=colorred1colorblue0colorred2colorblue0colorred3colorblue7colorred4colorblue6.colorred3colorblue9colorred4colorblue8colorred4colorblue7colorred9colorblue6colorred9colorblue9colorred9colorblue8colorred9colorblue7colorred9colorblue9colorred9colorblue8ldots$$
This is just one way. By the way, the name alternating decimal expansion is something I just made up, so don't take it too seriously.
Edited to add: I see you changed your criteria in a comment. This mapping is an injection, but not a surjection, so it doesn't have an inverse. You can make it surjective, but it's messy, and I'm not going to do it here.
Edited to add: As the OP points out in a comment, I have not define the mapping for $b,cle 0$. My bad! But you can fix this by using the decimal expansions of $e^b$ and $e^c$, which are always strictly greater than zero.
answered 2 days ago
TonyKTonyK
43k356135
$endgroup$
It is possible, but it's not pretty, and as VENKITESH says, it's certainly not linear. Here is one way: define $$f(a,b,c)=(x,y)$$ where $x=a$, and $y$ is the alternating decimal expansion of $b$ and $c$. What that means is this:
Take the non-terminating decimal expansions of $b$ and $c$; if necessary, insert leading zeroes so that the integral parts have the same length. For instance, if $b=1234.345$ and $c=76.987698798ldots$, we convert $b$ to its non-terminating form $b=colorred1234.344999999ldots$, and we write $c$ as $colorblue0076.987698798ldots$ to make the lengths before the decimal point equal.
Now we can write $y$ by taking digits from $b$ and $c$ in turn:
$$y=colorred1colorblue0colorred2colorblue0colorred3colorblue7colorred4colorblue6.colorred3colorblue9colorred4colorblue8colorred4colorblue7colorred9colorblue6colorred9colorblue9colorred9colorblue8colorred9colorblue7colorred9colorblue9colorred9colorblue8ldots$$
This is just one way. By the way, the name alternating decimal expansion is something I just made up, so don't take it too seriously.
Edited to add: I see you changed your criteria in a comment. This mapping is an injection, but not a surjection, so it doesn't have an inverse. You can make it surjective, but it's messy, and I'm not going to do it here.
Edited to add: As the OP points out in a comment, I have not define the mapping for $b,cle 0$. My bad! But you can fix this by using the decimal expansions of $e^b$ and $e^c$, which are always strictly greater than zero.
answered 2 days ago
TonyKTonyK
43k356135
edited 2 days ago
answered 2 days ago
TonyKTonyK
43k356135
answered 2 days ago
TonyKTonyK
43k356135
answered 2 days ago
TonyKTonyK
43k356135
43k356135
$begingroup$
(1,2,1)=(1,21) but (1,-2,-1) and (1,-2,1)=(?) Thanks for the example.
$endgroup$
– BJKShah
2 days ago
add a comment |
$begingroup$
(1,2,1)=(1,21) but (1,-2,-1) and (1,-2,1)=(?) Thanks for the example.
$endgroup$
– BJKShah
2 days ago
$begingroup$
(1,2,1)=(1,21) but (1,-2,-1) and (1,-2,1)=(?) Thanks for the example.
$endgroup$
– BJKShah
2 days ago
$begingroup$
(1,2,1)=(1,21) but (1,-2,-1) and (1,-2,1)=(?) Thanks for the example.
$endgroup$
– BJKShah
2 days ago
add a comment |
$begingroup$
It's is possible to have a bijection between $mathbbR^3$ and $mathbbR^2$, since it can be shown that $mathbbR$ and $mathbbR^2$, and hence every $mathbbR^n$ has the same cardinality. But a linear or affine isomorphism is not possible if the dimensions are different.
edited 2 days ago
tatan
5,79762760
answered 2 days ago
VENKITESHVENKITESH
36817
$endgroup$
$begingroup$
Can you please explain the last sentence a bit more?
$endgroup$
– BJKShah
2 days ago
$begingroup$
Right. A map f:R^nto R^m will be linear if f(u+v)=f(u)+f(v) and f(au)=af(u) for all scalars a and vectors u,v. We can show that if such an f is a bijection, then m=n. This follows from what is called the Rank Nullity Theorem. An affine map is of the form f+c, for a linear map f and a constant vector c. Again if an affine map is bijective, then we can show that m=n.
$endgroup$
– VENKITESH
2 days ago
$begingroup$
Any good books for studying this?
$endgroup$
– BJKShah
2 days ago
$begingroup$
Calculus and Analytic Geometry by Thomas and Finney, for coordinate geometry, and Linear Algebra by Gilbert Strang for linear algebra.
$endgroup$
– VENKITESH
2 days ago
add a comment |
$begingroup$
It's is possible to have a bijection between $mathbbR^3$ and $mathbbR^2$, since it can be shown that $mathbbR$ and $mathbbR^2$, and hence every $mathbbR^n$ has the same cardinality. But a linear or affine isomorphism is not possible if the dimensions are different.
edited 2 days ago
tatan
5,79762760
answered 2 days ago
VENKITESHVENKITESH
36817
$endgroup$
$begingroup$
Can you please explain the last sentence a bit more?
$endgroup$
– BJKShah
2 days ago
$begingroup$
Right. A map f:R^nto R^m will be linear if f(u+v)=f(u)+f(v) and f(au)=af(u) for all scalars a and vectors u,v. We can show that if such an f is a bijection, then m=n. This follows from what is called the Rank Nullity Theorem. An affine map is of the form f+c, for a linear map f and a constant vector c. Again if an affine map is bijective, then we can show that m=n.
$endgroup$
– VENKITESH
2 days ago
$begingroup$
Any good books for studying this?
$endgroup$
– BJKShah
2 days ago
$begingroup$
Calculus and Analytic Geometry by Thomas and Finney, for coordinate geometry, and Linear Algebra by Gilbert Strang for linear algebra.
$endgroup$
– VENKITESH
2 days ago
add a comment |
$begingroup$
It's is possible to have a bijection between $mathbbR^3$ and $mathbbR^2$, since it can be shown that $mathbbR$ and $mathbbR^2$, and hence every $mathbbR^n$ has the same cardinality. But a linear or affine isomorphism is not possible if the dimensions are different.
edited 2 days ago
tatan
5,79762760
answered 2 days ago
VENKITESHVENKITESH
36817
$endgroup$
It's is possible to have a bijection between $mathbbR^3$ and $mathbbR^2$, since it can be shown that $mathbbR$ and $mathbbR^2$, and hence every $mathbbR^n$ has the same cardinality. But a linear or affine isomorphism is not possible if the dimensions are different.
edited 2 days ago
tatan
5,79762760
answered 2 days ago
VENKITESHVENKITESH
36817
edited 2 days ago
tatan
5,79762760
edited 2 days ago
tatan
5,79762760
edited 2 days ago
tatan
5,79762760
5,79762760
answered 2 days ago
VENKITESHVENKITESH
36817
answered 2 days ago
VENKITESHVENKITESH
36817
answered 2 days ago
VENKITESHVENKITESH
36817
36817
$begingroup$
Can you please explain the last sentence a bit more?
$endgroup$
– BJKShah
2 days ago
$begingroup$
Right. A map f:R^nto R^m will be linear if f(u+v)=f(u)+f(v) and f(au)=af(u) for all scalars a and vectors u,v. We can show that if such an f is a bijection, then m=n. This follows from what is called the Rank Nullity Theorem. An affine map is of the form f+c, for a linear map f and a constant vector c. Again if an affine map is bijective, then we can show that m=n.
$endgroup$
– VENKITESH
2 days ago
$begingroup$
Any good books for studying this?
$endgroup$
– BJKShah
2 days ago
$begingroup$
Calculus and Analytic Geometry by Thomas and Finney, for coordinate geometry, and Linear Algebra by Gilbert Strang for linear algebra.
$endgroup$
– VENKITESH
2 days ago
add a comment |
$begingroup$
Can you please explain the last sentence a bit more?
$endgroup$
– BJKShah
2 days ago
$begingroup$
Right. A map f:R^nto R^m will be linear if f(u+v)=f(u)+f(v) and f(au)=af(u) for all scalars a and vectors u,v. We can show that if such an f is a bijection, then m=n. This follows from what is called the Rank Nullity Theorem. An affine map is of the form f+c, for a linear map f and a constant vector c. Again if an affine map is bijective, then we can show that m=n.
$endgroup$
– VENKITESH
2 days ago
$begingroup$
Any good books for studying this?
$endgroup$
– BJKShah
2 days ago
$begingroup$
Calculus and Analytic Geometry by Thomas and Finney, for coordinate geometry, and Linear Algebra by Gilbert Strang for linear algebra.
$endgroup$
– VENKITESH
2 days ago
$begingroup$
Can you please explain the last sentence a bit more?
$endgroup$
– BJKShah
2 days ago
$begingroup$
Can you please explain the last sentence a bit more?
$endgroup$
– BJKShah
2 days ago
$begingroup$
Right. A map f:R^nto R^m will be linear if f(u+v)=f(u)+f(v) and f(au)=af(u) for all scalars a and vectors u,v. We can show that if such an f is a bijection, then m=n. This follows from what is called the Rank Nullity Theorem. An affine map is of the form f+c, for a linear map f and a constant vector c. Again if an affine map is bijective, then we can show that m=n.
$endgroup$
– VENKITESH
2 days ago
$begingroup$
Right. A map f:R^nto R^m will be linear if f(u+v)=f(u)+f(v) and f(au)=af(u) for all scalars a and vectors u,v. We can show that if such an f is a bijection, then m=n. This follows from what is called the Rank Nullity Theorem. An affine map is of the form f+c, for a linear map f and a constant vector c. Again if an affine map is bijective, then we can show that m=n.
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– VENKITESH
2 days ago
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Any good books for studying this?
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– BJKShah
2 days ago
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Any good books for studying this?
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– BJKShah
2 days ago
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Calculus and Analytic Geometry by Thomas and Finney, for coordinate geometry, and Linear Algebra by Gilbert Strang for linear algebra.
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– VENKITESH
2 days ago
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Calculus and Analytic Geometry by Thomas and Finney, for coordinate geometry, and Linear Algebra by Gilbert Strang for linear algebra.
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– VENKITESH
2 days ago
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Just to clarify, do you want the map from $Bbb R^3$ to $Bbb R^2$ to be invertible, or just exist? As BJKShah, the former is possible but not if it's linear, whereas the latter is possible even then (such a map is an example of a projection).
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– J.G.
2 days ago
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I want the conversion to be possible in both directions.( How to convert former as you said , any links?) Thanks
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– BJKShah
2 days ago