Obtaining the gradient of a vector-valued function
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I have read that obtaining the gradient of a vector-valued function $f:mathbb{R}^n to mathbb{R}^m$ is the same as obtaining the Jacobian of this function.
Nevertheless, this function has only one argument (the vector $mathbf{x} in mathbb{R}^n$)
How can I take the gradient of a function $F(mathbf{x}_1, mathbf{x}_2, ldots, mathbf{x}_n)$ with respect to some $mathbf{x}_i$?
derivatives vector-spaces gradient-descent
asked Nov 15 at 8:22
The Bosco
495211
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up vote
0
down vote
favorite
I have read that obtaining the gradient of a vector-valued function $f:mathbb{R}^n to mathbb{R}^m$ is the same as obtaining the Jacobian of this function.
Nevertheless, this function has only one argument (the vector $mathbf{x} in mathbb{R}^n$)
How can I take the gradient of a function $F(mathbf{x}_1, mathbf{x}_2, ldots, mathbf{x}_n)$ with respect to some $mathbf{x}_i$?
derivatives vector-spaces gradient-descent
asked Nov 15 at 8:22
The Bosco
495211
A gradient of something with $n$ input indices and $m$ output indices will have $ntimes m$ indices. In this sense it is a tensor outer product. For example 1-tensor in like for example $3times 1$ vector, 1-tensor out $3times 1$ vector. 1+1 index = 2 indices. So your output becomes a 2-tensor (="matrix") (this is common for 3D vector field functions).
– mathreadler
Nov 15 at 9:01
Idk if it is correct to call it "gradient". It is called, instead, the Jacobian matrix of $f$ at a point. Yes, to obtain the Jacobian matrix respect to one of the variables just assume all the others are constants
– Masacroso
Nov 15 at 9:29
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have read that obtaining the gradient of a vector-valued function $f:mathbb{R}^n to mathbb{R}^m$ is the same as obtaining the Jacobian of this function.
Nevertheless, this function has only one argument (the vector $mathbf{x} in mathbb{R}^n$)
How can I take the gradient of a function $F(mathbf{x}_1, mathbf{x}_2, ldots, mathbf{x}_n)$ with respect to some $mathbf{x}_i$?
derivatives vector-spaces gradient-descent
asked Nov 15 at 8:22
The Bosco
495211
I have read that obtaining the gradient of a vector-valued function $f:mathbb{R}^n to mathbb{R}^m$ is the same as obtaining the Jacobian of this function.
Nevertheless, this function has only one argument (the vector $mathbf{x} in mathbb{R}^n$)
How can I take the gradient of a function $F(mathbf{x}_1, mathbf{x}_2, ldots, mathbf{x}_n)$ with respect to some $mathbf{x}_i$?
derivatives vector-spaces gradient-descent
derivatives vector-spaces gradient-descent
asked Nov 15 at 8:22
The Bosco
495211
asked Nov 15 at 8:22
The Bosco
495211
asked Nov 15 at 8:22
The Bosco
495211
asked Nov 15 at 8:22
The Bosco
495211
asked Nov 15 at 8:22
The Bosco
495211
495211
A gradient of something with $n$ input indices and $m$ output indices will have $ntimes m$ indices. In this sense it is a tensor outer product. For example 1-tensor in like for example $3times 1$ vector, 1-tensor out $3times 1$ vector. 1+1 index = 2 indices. So your output becomes a 2-tensor (="matrix") (this is common for 3D vector field functions).
– mathreadler
Nov 15 at 9:01
Idk if it is correct to call it "gradient". It is called, instead, the Jacobian matrix of $f$ at a point. Yes, to obtain the Jacobian matrix respect to one of the variables just assume all the others are constants
– Masacroso
Nov 15 at 9:29
add a comment |
A gradient of something with $n$ input indices and $m$ output indices will have $ntimes m$ indices. In this sense it is a tensor outer product. For example 1-tensor in like for example $3times 1$ vector, 1-tensor out $3times 1$ vector. 1+1 index = 2 indices. So your output becomes a 2-tensor (="matrix") (this is common for 3D vector field functions).
– mathreadler
Nov 15 at 9:01
Idk if it is correct to call it "gradient". It is called, instead, the Jacobian matrix of $f$ at a point. Yes, to obtain the Jacobian matrix respect to one of the variables just assume all the others are constants
– Masacroso
Nov 15 at 9:29
A gradient of something with $n$ input indices and $m$ output indices will have $ntimes m$ indices. In this sense it is a tensor outer product. For example 1-tensor in like for example $3times 1$ vector, 1-tensor out $3times 1$ vector. 1+1 index = 2 indices. So your output becomes a 2-tensor (="matrix") (this is common for 3D vector field functions).
– mathreadler
Nov 15 at 9:01
A gradient of something with $n$ input indices and $m$ output indices will have $ntimes m$ indices. In this sense it is a tensor outer product. For example 1-tensor in like for example $3times 1$ vector, 1-tensor out $3times 1$ vector. 1+1 index = 2 indices. So your output becomes a 2-tensor (="matrix") (this is common for 3D vector field functions).
– mathreadler
Nov 15 at 9:01
Idk if it is correct to call it "gradient". It is called, instead, the Jacobian matrix of $f$ at a point. Yes, to obtain the Jacobian matrix respect to one of the variables just assume all the others are constants
– Masacroso
Nov 15 at 9:29
Idk if it is correct to call it "gradient". It is called, instead, the Jacobian matrix of $f$ at a point. Yes, to obtain the Jacobian matrix respect to one of the variables just assume all the others are constants
– Masacroso
Nov 15 at 9:29
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
The 'gradient' of something usually means taking all partial derivatives. Therefore
"taking the gradient of a function $F(x_1,x_2,....,x_n)$ with respect to some $x_i$"
is not really a thing. However you are right that all partial derivatives of a vector valued function arranged in a $mtimes n$ matrix are usually referd to as Jacobian.
answered Nov 15 at 8:30
maxmilgram
4227
This is a homework question. For some specific function $F$ I have to do that.
– The Bosco
Nov 15 at 8:33
The question is to calculate the gradient? Then just calculate all partial derivatives. If that doesn't help, please provide the full problem description.
– maxmilgram
Nov 15 at 8:34
add a comment |
up vote
0
down vote
A gradient of something with $n$ input indices and $m$ output indices will have $ntimes m$ indices. In this sense it is a tensor outer product. For example $1-$tensor in like for example $3times 1$ vector, $1-$tensor out $3times 1$ vector.
Since $1+1 = 2$, your output becomes a 2-tensor (="matrix") (this is common for 3D vector field functions).
answered Nov 15 at 9:08
mathreadler
14.6k72160
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The 'gradient' of something usually means taking all partial derivatives. Therefore
"taking the gradient of a function $F(x_1,x_2,....,x_n)$ with respect to some $x_i$"
is not really a thing. However you are right that all partial derivatives of a vector valued function arranged in a $mtimes n$ matrix are usually referd to as Jacobian.
answered Nov 15 at 8:30
maxmilgram
4227
This is a homework question. For some specific function $F$ I have to do that.
– The Bosco
Nov 15 at 8:33
The question is to calculate the gradient? Then just calculate all partial derivatives. If that doesn't help, please provide the full problem description.
– maxmilgram
Nov 15 at 8:34
add a comment |
up vote
0
down vote
The 'gradient' of something usually means taking all partial derivatives. Therefore
"taking the gradient of a function $F(x_1,x_2,....,x_n)$ with respect to some $x_i$"
is not really a thing. However you are right that all partial derivatives of a vector valued function arranged in a $mtimes n$ matrix are usually referd to as Jacobian.
answered Nov 15 at 8:30
maxmilgram
4227
This is a homework question. For some specific function $F$ I have to do that.
– The Bosco
Nov 15 at 8:33
The question is to calculate the gradient? Then just calculate all partial derivatives. If that doesn't help, please provide the full problem description.
– maxmilgram
Nov 15 at 8:34
add a comment |
up vote
0
down vote
up vote
0
down vote
The 'gradient' of something usually means taking all partial derivatives. Therefore
"taking the gradient of a function $F(x_1,x_2,....,x_n)$ with respect to some $x_i$"
is not really a thing. However you are right that all partial derivatives of a vector valued function arranged in a $mtimes n$ matrix are usually referd to as Jacobian.
answered Nov 15 at 8:30
maxmilgram
4227
The 'gradient' of something usually means taking all partial derivatives. Therefore
"taking the gradient of a function $F(x_1,x_2,....,x_n)$ with respect to some $x_i$"
is not really a thing. However you are right that all partial derivatives of a vector valued function arranged in a $mtimes n$ matrix are usually referd to as Jacobian.
answered Nov 15 at 8:30
maxmilgram
4227
answered Nov 15 at 8:30
maxmilgram
4227
answered Nov 15 at 8:30
maxmilgram
4227
answered Nov 15 at 8:30
maxmilgram
4227
4227
This is a homework question. For some specific function $F$ I have to do that.
– The Bosco
Nov 15 at 8:33
The question is to calculate the gradient? Then just calculate all partial derivatives. If that doesn't help, please provide the full problem description.
– maxmilgram
Nov 15 at 8:34
add a comment |
This is a homework question. For some specific function $F$ I have to do that.
– The Bosco
Nov 15 at 8:33
The question is to calculate the gradient? Then just calculate all partial derivatives. If that doesn't help, please provide the full problem description.
– maxmilgram
Nov 15 at 8:34
This is a homework question. For some specific function $F$ I have to do that.
– The Bosco
Nov 15 at 8:33
This is a homework question. For some specific function $F$ I have to do that.
– The Bosco
Nov 15 at 8:33
The question is to calculate the gradient? Then just calculate all partial derivatives. If that doesn't help, please provide the full problem description.
– maxmilgram
Nov 15 at 8:34
The question is to calculate the gradient? Then just calculate all partial derivatives. If that doesn't help, please provide the full problem description.
– maxmilgram
Nov 15 at 8:34
add a comment |
up vote
0
down vote
A gradient of something with $n$ input indices and $m$ output indices will have $ntimes m$ indices. In this sense it is a tensor outer product. For example $1-$tensor in like for example $3times 1$ vector, $1-$tensor out $3times 1$ vector.
Since $1+1 = 2$, your output becomes a 2-tensor (="matrix") (this is common for 3D vector field functions).
answered Nov 15 at 9:08
mathreadler
14.6k72160
add a comment |
up vote
0
down vote
A gradient of something with $n$ input indices and $m$ output indices will have $ntimes m$ indices. In this sense it is a tensor outer product. For example $1-$tensor in like for example $3times 1$ vector, $1-$tensor out $3times 1$ vector.
Since $1+1 = 2$, your output becomes a 2-tensor (="matrix") (this is common for 3D vector field functions).
answered Nov 15 at 9:08
mathreadler
14.6k72160
add a comment |
up vote
0
down vote
up vote
0
down vote
A gradient of something with $n$ input indices and $m$ output indices will have $ntimes m$ indices. In this sense it is a tensor outer product. For example $1-$tensor in like for example $3times 1$ vector, $1-$tensor out $3times 1$ vector.
Since $1+1 = 2$, your output becomes a 2-tensor (="matrix") (this is common for 3D vector field functions).
answered Nov 15 at 9:08
mathreadler
14.6k72160
A gradient of something with $n$ input indices and $m$ output indices will have $ntimes m$ indices. In this sense it is a tensor outer product. For example $1-$tensor in like for example $3times 1$ vector, $1-$tensor out $3times 1$ vector.
Since $1+1 = 2$, your output becomes a 2-tensor (="matrix") (this is common for 3D vector field functions).
answered Nov 15 at 9:08
mathreadler
14.6k72160
answered Nov 15 at 9:08
mathreadler
14.6k72160
answered Nov 15 at 9:08
mathreadler
14.6k72160
answered Nov 15 at 9:08
mathreadler
14.6k72160
14.6k72160
add a comment |
add a comment |
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A gradient of something with $n$ input indices and $m$ output indices will have $ntimes m$ indices. In this sense it is a tensor outer product. For example 1-tensor in like for example $3times 1$ vector, 1-tensor out $3times 1$ vector. 1+1 index = 2 indices. So your output becomes a 2-tensor (="matrix") (this is common for 3D vector field functions).
– mathreadler
Nov 15 at 9:01
Idk if it is correct to call it "gradient". It is called, instead, the Jacobian matrix of $f$ at a point. Yes, to obtain the Jacobian matrix respect to one of the variables just assume all the others are constants
– Masacroso
Nov 15 at 9:29