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Get solutions of an equation given its output
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Lets assume I have a function $f(x_1, x_2, x_3)$ that takes $x_1, x_2, x_3$ as inputs. For the sake of simplicity, lets assume that this $f(x)$ is a linear equation.
My question is- if I decide that the output of my function has to be an arbitrary value- $v$, can I generate all or a fixed number of the possible solutions or combinations of values that when passed to $f$ will generate $v$?
If yes, how do I achieve this?
linear-algebra
asked Jul 26 at 10:09
Clock Slave
432415
add a comment |Â
up vote
0
down vote
favorite
Lets assume I have a function $f(x_1, x_2, x_3)$ that takes $x_1, x_2, x_3$ as inputs. For the sake of simplicity, lets assume that this $f(x)$ is a linear equation.
My question is- if I decide that the output of my function has to be an arbitrary value- $v$, can I generate all or a fixed number of the possible solutions or combinations of values that when passed to $f$ will generate $v$?
If yes, how do I achieve this?
linear-algebra
asked Jul 26 at 10:09
Clock Slave
432415
What you're describing there is just the problem of solving a linear equation.
– Jack M
Jul 26 at 11:24
@ClockSlave Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Lets assume I have a function $f(x_1, x_2, x_3)$ that takes $x_1, x_2, x_3$ as inputs. For the sake of simplicity, lets assume that this $f(x)$ is a linear equation.
My question is- if I decide that the output of my function has to be an arbitrary value- $v$, can I generate all or a fixed number of the possible solutions or combinations of values that when passed to $f$ will generate $v$?
If yes, how do I achieve this?
linear-algebra
asked Jul 26 at 10:09
Clock Slave
432415
Lets assume I have a function $f(x_1, x_2, x_3)$ that takes $x_1, x_2, x_3$ as inputs. For the sake of simplicity, lets assume that this $f(x)$ is a linear equation.
My question is- if I decide that the output of my function has to be an arbitrary value- $v$, can I generate all or a fixed number of the possible solutions or combinations of values that when passed to $f$ will generate $v$?
If yes, how do I achieve this?
linear-algebra
asked Jul 26 at 10:09
Clock Slave
432415
asked Jul 26 at 10:09
Clock Slave
432415
asked Jul 26 at 10:09
Clock Slave
432415
asked Jul 26 at 10:09
Clock Slave
432415
432415
What you're describing there is just the problem of solving a linear equation.
– Jack M
Jul 26 at 11:24
@ClockSlave Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
add a comment |Â
What you're describing there is just the problem of solving a linear equation.
– Jack M
Jul 26 at 11:24
@ClockSlave Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
What you're describing there is just the problem of solving a linear equation.
– Jack M
Jul 26 at 11:24
What you're describing there is just the problem of solving a linear equation.
– Jack M
Jul 26 at 11:24
@ClockSlave Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
@ClockSlave Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
0
down vote
Since you assume that $f(x)$ is linear, it takes the form $a_0 + a_1x_1+a_2x_2+a_3x_3$. Let $v'=v-a_0.$ If $mathbfa=(a_1,a_2,a_3)$, you have to find all $mathbfxinmathbbR^3$ such that $mathbfa.mathbfx=v'$.
To find $mathbfx$, we can do a coordinate transformation so that $mathbfa$ is aligned with the $x$-axis and then $mathbfx$ is any vector whose $x$ component equals $v'$ in the new system. Use inverse coordinate transformation to find $mathbfx$ in the original coordinate system.
I'm sorry for the partial answer but I've forgotten how to do these steps. Maybe someone else can elaborate.
answered Jul 26 at 10:26
Shirish Kulhari
929215
add a comment |Â
up vote
0
down vote
The equation $f(x_1,x_2,x_3)=v$ describes a surface in the space $(x_1,x_2,x_3)$. This is called the implicit equation of the surface. If you vary $v$, you will get another surface "nested" in the first. For instance, $x_1^2+x_2^2+x_3^2=v$ gives you a family of concentric spheres of radius $sqrt v$.
In the case of a linear transform, these surfaces are parallel planes.
You can obtain all solutions of
$$ax_1+bx_2+cx_3=v$$ by setting two coordinates arbitrarily and computing the third one, e.g.
$$x_3=fracv-ax_1-bx_2c.$$
answered Jul 26 at 10:43
Yves Daoust
111k665203
Why are we assuming that $x_1, x_2$ and $x_3$ are scalars? It is not stated in the OP.
– gimusi
Jul 26 at 11:49
1
@gimusi: Why would we be assuming that $x_1$, $x_2$ and $x_3$ are vectors ? It is not stated in the OP.
– Yves Daoust
Jul 26 at 12:01
For example because the vector case is more general and the scalar case can be treated as a particular case.
– gimusi
Jul 26 at 12:03
@gimusi: then in your answer make sure to use matrices to express the linear function as this is more general and the dot product case can be treated as a particular case. Don't forget to specify that $A_1$ must be invertible.
– Yves Daoust
Jul 26 at 12:06
I don't understand this last comment, anyway I think that the answers given are sufficent for the OP. Bye
– gimusi
Jul 26 at 12:08
add a comment |Â
up vote
-1
down vote
Yes, for a linear function it is always possible to reconstruct $x$ from $v$ given that $v$ is a solution.
In general, for $f:mathbbR^ntomathbbR^m$ to do that we simply need to solve the linear system $Ax=v$, where $A$ is the matrix representation of $f$ for a given basis.
For the particular case of $f:mathbbR^3tomathbbR$, we have that $x_1$,$x_2$ and $x_3$ scalars and we need to solve
$$a_1x_1+a_2x_2+a_3x_3=v$$
which has infinitely many solutions, that is setting $x_2=s$ and $x_3=t$
$$x_1=fracv-a_2s-a_3 ta_1$$
answered Jul 26 at 10:21
gimusi
65k73583
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Since you assume that $f(x)$ is linear, it takes the form $a_0 + a_1x_1+a_2x_2+a_3x_3$. Let $v'=v-a_0.$ If $mathbfa=(a_1,a_2,a_3)$, you have to find all $mathbfxinmathbbR^3$ such that $mathbfa.mathbfx=v'$.
To find $mathbfx$, we can do a coordinate transformation so that $mathbfa$ is aligned with the $x$-axis and then $mathbfx$ is any vector whose $x$ component equals $v'$ in the new system. Use inverse coordinate transformation to find $mathbfx$ in the original coordinate system.
I'm sorry for the partial answer but I've forgotten how to do these steps. Maybe someone else can elaborate.
answered Jul 26 at 10:26
Shirish Kulhari
929215
add a comment |Â
up vote
0
down vote
Since you assume that $f(x)$ is linear, it takes the form $a_0 + a_1x_1+a_2x_2+a_3x_3$. Let $v'=v-a_0.$ If $mathbfa=(a_1,a_2,a_3)$, you have to find all $mathbfxinmathbbR^3$ such that $mathbfa.mathbfx=v'$.
To find $mathbfx$, we can do a coordinate transformation so that $mathbfa$ is aligned with the $x$-axis and then $mathbfx$ is any vector whose $x$ component equals $v'$ in the new system. Use inverse coordinate transformation to find $mathbfx$ in the original coordinate system.
I'm sorry for the partial answer but I've forgotten how to do these steps. Maybe someone else can elaborate.
answered Jul 26 at 10:26
Shirish Kulhari
929215
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since you assume that $f(x)$ is linear, it takes the form $a_0 + a_1x_1+a_2x_2+a_3x_3$. Let $v'=v-a_0.$ If $mathbfa=(a_1,a_2,a_3)$, you have to find all $mathbfxinmathbbR^3$ such that $mathbfa.mathbfx=v'$.
To find $mathbfx$, we can do a coordinate transformation so that $mathbfa$ is aligned with the $x$-axis and then $mathbfx$ is any vector whose $x$ component equals $v'$ in the new system. Use inverse coordinate transformation to find $mathbfx$ in the original coordinate system.
I'm sorry for the partial answer but I've forgotten how to do these steps. Maybe someone else can elaborate.
answered Jul 26 at 10:26
Shirish Kulhari
929215
Since you assume that $f(x)$ is linear, it takes the form $a_0 + a_1x_1+a_2x_2+a_3x_3$. Let $v'=v-a_0.$ If $mathbfa=(a_1,a_2,a_3)$, you have to find all $mathbfxinmathbbR^3$ such that $mathbfa.mathbfx=v'$.
To find $mathbfx$, we can do a coordinate transformation so that $mathbfa$ is aligned with the $x$-axis and then $mathbfx$ is any vector whose $x$ component equals $v'$ in the new system. Use inverse coordinate transformation to find $mathbfx$ in the original coordinate system.
I'm sorry for the partial answer but I've forgotten how to do these steps. Maybe someone else can elaborate.
answered Jul 26 at 10:26
Shirish Kulhari
929215
answered Jul 26 at 10:26
Shirish Kulhari
929215
answered Jul 26 at 10:26
Shirish Kulhari
929215
answered Jul 26 at 10:26
Shirish Kulhari
929215
929215
add a comment |Â
add a comment |Â
up vote
0
down vote
The equation $f(x_1,x_2,x_3)=v$ describes a surface in the space $(x_1,x_2,x_3)$. This is called the implicit equation of the surface. If you vary $v$, you will get another surface "nested" in the first. For instance, $x_1^2+x_2^2+x_3^2=v$ gives you a family of concentric spheres of radius $sqrt v$.
In the case of a linear transform, these surfaces are parallel planes.
You can obtain all solutions of
$$ax_1+bx_2+cx_3=v$$ by setting two coordinates arbitrarily and computing the third one, e.g.
$$x_3=fracv-ax_1-bx_2c.$$
answered Jul 26 at 10:43
Yves Daoust
111k665203
Why are we assuming that $x_1, x_2$ and $x_3$ are scalars? It is not stated in the OP.
– gimusi
Jul 26 at 11:49
1
@gimusi: Why would we be assuming that $x_1$, $x_2$ and $x_3$ are vectors ? It is not stated in the OP.
– Yves Daoust
Jul 26 at 12:01
For example because the vector case is more general and the scalar case can be treated as a particular case.
– gimusi
Jul 26 at 12:03
@gimusi: then in your answer make sure to use matrices to express the linear function as this is more general and the dot product case can be treated as a particular case. Don't forget to specify that $A_1$ must be invertible.
– Yves Daoust
Jul 26 at 12:06
I don't understand this last comment, anyway I think that the answers given are sufficent for the OP. Bye
– gimusi
Jul 26 at 12:08
add a comment |Â
up vote
0
down vote
The equation $f(x_1,x_2,x_3)=v$ describes a surface in the space $(x_1,x_2,x_3)$. This is called the implicit equation of the surface. If you vary $v$, you will get another surface "nested" in the first. For instance, $x_1^2+x_2^2+x_3^2=v$ gives you a family of concentric spheres of radius $sqrt v$.
In the case of a linear transform, these surfaces are parallel planes.
You can obtain all solutions of
$$ax_1+bx_2+cx_3=v$$ by setting two coordinates arbitrarily and computing the third one, e.g.
$$x_3=fracv-ax_1-bx_2c.$$
answered Jul 26 at 10:43
Yves Daoust
111k665203
Why are we assuming that $x_1, x_2$ and $x_3$ are scalars? It is not stated in the OP.
– gimusi
Jul 26 at 11:49
1
@gimusi: Why would we be assuming that $x_1$, $x_2$ and $x_3$ are vectors ? It is not stated in the OP.
– Yves Daoust
Jul 26 at 12:01
For example because the vector case is more general and the scalar case can be treated as a particular case.
– gimusi
Jul 26 at 12:03
@gimusi: then in your answer make sure to use matrices to express the linear function as this is more general and the dot product case can be treated as a particular case. Don't forget to specify that $A_1$ must be invertible.
– Yves Daoust
Jul 26 at 12:06
I don't understand this last comment, anyway I think that the answers given are sufficent for the OP. Bye
– gimusi
Jul 26 at 12:08
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The equation $f(x_1,x_2,x_3)=v$ describes a surface in the space $(x_1,x_2,x_3)$. This is called the implicit equation of the surface. If you vary $v$, you will get another surface "nested" in the first. For instance, $x_1^2+x_2^2+x_3^2=v$ gives you a family of concentric spheres of radius $sqrt v$.
In the case of a linear transform, these surfaces are parallel planes.
You can obtain all solutions of
$$ax_1+bx_2+cx_3=v$$ by setting two coordinates arbitrarily and computing the third one, e.g.
$$x_3=fracv-ax_1-bx_2c.$$
answered Jul 26 at 10:43
Yves Daoust
111k665203
The equation $f(x_1,x_2,x_3)=v$ describes a surface in the space $(x_1,x_2,x_3)$. This is called the implicit equation of the surface. If you vary $v$, you will get another surface "nested" in the first. For instance, $x_1^2+x_2^2+x_3^2=v$ gives you a family of concentric spheres of radius $sqrt v$.
In the case of a linear transform, these surfaces are parallel planes.
You can obtain all solutions of
$$ax_1+bx_2+cx_3=v$$ by setting two coordinates arbitrarily and computing the third one, e.g.
$$x_3=fracv-ax_1-bx_2c.$$
answered Jul 26 at 10:43
Yves Daoust
111k665203
answered Jul 26 at 10:43
Yves Daoust
111k665203
answered Jul 26 at 10:43
Yves Daoust
111k665203
answered Jul 26 at 10:43
Yves Daoust
111k665203
111k665203
Why are we assuming that $x_1, x_2$ and $x_3$ are scalars? It is not stated in the OP.
– gimusi
Jul 26 at 11:49
1
@gimusi: Why would we be assuming that $x_1$, $x_2$ and $x_3$ are vectors ? It is not stated in the OP.
– Yves Daoust
Jul 26 at 12:01
For example because the vector case is more general and the scalar case can be treated as a particular case.
– gimusi
Jul 26 at 12:03
@gimusi: then in your answer make sure to use matrices to express the linear function as this is more general and the dot product case can be treated as a particular case. Don't forget to specify that $A_1$ must be invertible.
– Yves Daoust
Jul 26 at 12:06
I don't understand this last comment, anyway I think that the answers given are sufficent for the OP. Bye
– gimusi
Jul 26 at 12:08
add a comment |Â
Why are we assuming that $x_1, x_2$ and $x_3$ are scalars? It is not stated in the OP.
– gimusi
Jul 26 at 11:49
1
@gimusi: Why would we be assuming that $x_1$, $x_2$ and $x_3$ are vectors ? It is not stated in the OP.
– Yves Daoust
Jul 26 at 12:01
For example because the vector case is more general and the scalar case can be treated as a particular case.
– gimusi
Jul 26 at 12:03
@gimusi: then in your answer make sure to use matrices to express the linear function as this is more general and the dot product case can be treated as a particular case. Don't forget to specify that $A_1$ must be invertible.
– Yves Daoust
Jul 26 at 12:06
I don't understand this last comment, anyway I think that the answers given are sufficent for the OP. Bye
– gimusi
Jul 26 at 12:08
Why are we assuming that $x_1, x_2$ and $x_3$ are scalars? It is not stated in the OP.
– gimusi
Jul 26 at 11:49
Why are we assuming that $x_1, x_2$ and $x_3$ are scalars? It is not stated in the OP.
– gimusi
Jul 26 at 11:49
1
1
@gimusi: Why would we be assuming that $x_1$, $x_2$ and $x_3$ are vectors ? It is not stated in the OP.
– Yves Daoust
Jul 26 at 12:01
@gimusi: Why would we be assuming that $x_1$, $x_2$ and $x_3$ are vectors ? It is not stated in the OP.
– Yves Daoust
Jul 26 at 12:01
For example because the vector case is more general and the scalar case can be treated as a particular case.
– gimusi
Jul 26 at 12:03
For example because the vector case is more general and the scalar case can be treated as a particular case.
– gimusi
Jul 26 at 12:03
@gimusi: then in your answer make sure to use matrices to express the linear function as this is more general and the dot product case can be treated as a particular case. Don't forget to specify that $A_1$ must be invertible.
– Yves Daoust
Jul 26 at 12:06
@gimusi: then in your answer make sure to use matrices to express the linear function as this is more general and the dot product case can be treated as a particular case. Don't forget to specify that $A_1$ must be invertible.
– Yves Daoust
Jul 26 at 12:06
I don't understand this last comment, anyway I think that the answers given are sufficent for the OP. Bye
– gimusi
Jul 26 at 12:08
I don't understand this last comment, anyway I think that the answers given are sufficent for the OP. Bye
– gimusi
Jul 26 at 12:08
add a comment |Â
up vote
-1
down vote
Yes, for a linear function it is always possible to reconstruct $x$ from $v$ given that $v$ is a solution.
In general, for $f:mathbbR^ntomathbbR^m$ to do that we simply need to solve the linear system $Ax=v$, where $A$ is the matrix representation of $f$ for a given basis.
For the particular case of $f:mathbbR^3tomathbbR$, we have that $x_1$,$x_2$ and $x_3$ scalars and we need to solve
$$a_1x_1+a_2x_2+a_3x_3=v$$
which has infinitely many solutions, that is setting $x_2=s$ and $x_3=t$
$$x_1=fracv-a_2s-a_3 ta_1$$
answered Jul 26 at 10:21
gimusi
65k73583
add a comment |Â
up vote
-1
down vote
Yes, for a linear function it is always possible to reconstruct $x$ from $v$ given that $v$ is a solution.
In general, for $f:mathbbR^ntomathbbR^m$ to do that we simply need to solve the linear system $Ax=v$, where $A$ is the matrix representation of $f$ for a given basis.
For the particular case of $f:mathbbR^3tomathbbR$, we have that $x_1$,$x_2$ and $x_3$ scalars and we need to solve
$$a_1x_1+a_2x_2+a_3x_3=v$$
which has infinitely many solutions, that is setting $x_2=s$ and $x_3=t$
$$x_1=fracv-a_2s-a_3 ta_1$$
answered Jul 26 at 10:21
gimusi
65k73583
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
Yes, for a linear function it is always possible to reconstruct $x$ from $v$ given that $v$ is a solution.
In general, for $f:mathbbR^ntomathbbR^m$ to do that we simply need to solve the linear system $Ax=v$, where $A$ is the matrix representation of $f$ for a given basis.
For the particular case of $f:mathbbR^3tomathbbR$, we have that $x_1$,$x_2$ and $x_3$ scalars and we need to solve
$$a_1x_1+a_2x_2+a_3x_3=v$$
which has infinitely many solutions, that is setting $x_2=s$ and $x_3=t$
$$x_1=fracv-a_2s-a_3 ta_1$$
answered Jul 26 at 10:21
gimusi
65k73583
Yes, for a linear function it is always possible to reconstruct $x$ from $v$ given that $v$ is a solution.
In general, for $f:mathbbR^ntomathbbR^m$ to do that we simply need to solve the linear system $Ax=v$, where $A$ is the matrix representation of $f$ for a given basis.
For the particular case of $f:mathbbR^3tomathbbR$, we have that $x_1$,$x_2$ and $x_3$ scalars and we need to solve
$$a_1x_1+a_2x_2+a_3x_3=v$$
which has infinitely many solutions, that is setting $x_2=s$ and $x_3=t$
$$x_1=fracv-a_2s-a_3 ta_1$$
answered Jul 26 at 10:21
gimusi
65k73583
edited Jul 26 at 12:03
answered Jul 26 at 10:21
gimusi
65k73583
answered Jul 26 at 10:21
gimusi
65k73583
answered Jul 26 at 10:21
gimusi
65k73583
65k73583
add a comment |Â
add a comment |Â
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What you're describing there is just the problem of solving a linear equation.
– Jack M
Jul 26 at 11:24
@ClockSlave Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday