Hessian-vector products
$begingroup$
Can someone explain why this is true?
$$g(x + Delta x) = g(x) + H(x) Delta x$$
where g is the gradient of function f(x) with respect to x, and H is the hessian of f(x) with respect to x.
I would really appreciate a detailed derivation because I don't understand what it means to take the gradient/derivative of $Delta x$ which represents the change in x.
Source towards the top here: https://justindomke.wordpress.com/2009/01/17/hessian-vector-products/
optimization machine-learning
asked Nov 28 '18 at 6:41
user3180user3180
1185
$endgroup$
|
show 1 more comment
$begingroup$
Can someone explain why this is true?
$$g(x + Delta x) = g(x) + H(x) Delta x$$
where g is the gradient of function f(x) with respect to x, and H is the hessian of f(x) with respect to x.
I would really appreciate a detailed derivation because I don't understand what it means to take the gradient/derivative of $Delta x$ which represents the change in x.
Source towards the top here: https://justindomke.wordpress.com/2009/01/17/hessian-vector-products/
optimization machine-learning
asked Nov 28 '18 at 6:41
user3180user3180
1185
$endgroup$
$begingroup$
Is it like the third term in the taylor expansion? en.m.wikipedia.org/wiki/…
$endgroup$
– Emil
Nov 28 '18 at 7:14
$begingroup$
Is the lhs supposed to be the gradient taken at a point or is it a scalar? The hessian is a second order tensor, so it should eat two dx to become a scalar, not one, I believe.
$endgroup$
– Emil
Nov 28 '18 at 7:22
$begingroup$
$Delta x$ is a vector I believe
$endgroup$
– user3180
Nov 28 '18 at 7:58
$begingroup$
I don't know why you're talking about scalars, the Hessian H(x) is a vector and so is $Delta x$. Is it because the source link is $g(x) = frac{d}{dx} f(x)$ is using scalar notation? He intends this to be the gradient of a vector
$endgroup$
– user3180
Nov 28 '18 at 8:01
$begingroup$
user3180: to know if there is an error and if so if it is on lhs or rhs (hessian is a second order tensor not a vector, according to some definitions, or maybe even third order tensor)
$endgroup$
– Emil
Nov 28 '18 at 13:26
|
show 1 more comment
$begingroup$
Can someone explain why this is true?
$$g(x + Delta x) = g(x) + H(x) Delta x$$
where g is the gradient of function f(x) with respect to x, and H is the hessian of f(x) with respect to x.
I would really appreciate a detailed derivation because I don't understand what it means to take the gradient/derivative of $Delta x$ which represents the change in x.
Source towards the top here: https://justindomke.wordpress.com/2009/01/17/hessian-vector-products/
optimization machine-learning
asked Nov 28 '18 at 6:41
user3180user3180
1185
$endgroup$
Can someone explain why this is true?
$$g(x + Delta x) = g(x) + H(x) Delta x$$
where g is the gradient of function f(x) with respect to x, and H is the hessian of f(x) with respect to x.
I would really appreciate a detailed derivation because I don't understand what it means to take the gradient/derivative of $Delta x$ which represents the change in x.
Source towards the top here: https://justindomke.wordpress.com/2009/01/17/hessian-vector-products/
optimization machine-learning
optimization machine-learning
asked Nov 28 '18 at 6:41
user3180user3180
1185
asked Nov 28 '18 at 6:41
user3180user3180
1185
edited Nov 28 '18 at 20:48
user3180
asked Nov 28 '18 at 6:41
user3180user3180
1185
asked Nov 28 '18 at 6:41
user3180user3180
1185
asked Nov 28 '18 at 6:41
user3180user3180
1185
1185
$begingroup$
Is it like the third term in the taylor expansion? en.m.wikipedia.org/wiki/…
$endgroup$
– Emil
Nov 28 '18 at 7:14
$begingroup$
Is the lhs supposed to be the gradient taken at a point or is it a scalar? The hessian is a second order tensor, so it should eat two dx to become a scalar, not one, I believe.
$endgroup$
– Emil
Nov 28 '18 at 7:22
$begingroup$
$Delta x$ is a vector I believe
$endgroup$
– user3180
Nov 28 '18 at 7:58
$begingroup$
I don't know why you're talking about scalars, the Hessian H(x) is a vector and so is $Delta x$. Is it because the source link is $g(x) = frac{d}{dx} f(x)$ is using scalar notation? He intends this to be the gradient of a vector
$endgroup$
– user3180
Nov 28 '18 at 8:01
$begingroup$
user3180: to know if there is an error and if so if it is on lhs or rhs (hessian is a second order tensor not a vector, according to some definitions, or maybe even third order tensor)
$endgroup$
– Emil
Nov 28 '18 at 13:26
|
show 1 more comment
$begingroup$
Is it like the third term in the taylor expansion? en.m.wikipedia.org/wiki/…
$endgroup$
– Emil
Nov 28 '18 at 7:14
$begingroup$
Is the lhs supposed to be the gradient taken at a point or is it a scalar? The hessian is a second order tensor, so it should eat two dx to become a scalar, not one, I believe.
$endgroup$
– Emil
Nov 28 '18 at 7:22
$begingroup$
$Delta x$ is a vector I believe
$endgroup$
– user3180
Nov 28 '18 at 7:58
$begingroup$
I don't know why you're talking about scalars, the Hessian H(x) is a vector and so is $Delta x$. Is it because the source link is $g(x) = frac{d}{dx} f(x)$ is using scalar notation? He intends this to be the gradient of a vector
$endgroup$
– user3180
Nov 28 '18 at 8:01
$begingroup$
user3180: to know if there is an error and if so if it is on lhs or rhs (hessian is a second order tensor not a vector, according to some definitions, or maybe even third order tensor)
$endgroup$
– Emil
Nov 28 '18 at 13:26
$begingroup$
Is it like the third term in the taylor expansion? en.m.wikipedia.org/wiki/…
$endgroup$
– Emil
Nov 28 '18 at 7:14
$begingroup$
Is it like the third term in the taylor expansion? en.m.wikipedia.org/wiki/…
$endgroup$
– Emil
Nov 28 '18 at 7:14
$begingroup$
Is the lhs supposed to be the gradient taken at a point or is it a scalar? The hessian is a second order tensor, so it should eat two dx to become a scalar, not one, I believe.
$endgroup$
– Emil
Nov 28 '18 at 7:22
$begingroup$
Is the lhs supposed to be the gradient taken at a point or is it a scalar? The hessian is a second order tensor, so it should eat two dx to become a scalar, not one, I believe.
$endgroup$
– Emil
Nov 28 '18 at 7:22
$begingroup$
$Delta x$ is a vector I believe
$endgroup$
– user3180
Nov 28 '18 at 7:58
$begingroup$
$Delta x$ is a vector I believe
$endgroup$
– user3180
Nov 28 '18 at 7:58
$begingroup$
I don't know why you're talking about scalars, the Hessian H(x) is a vector and so is $Delta x$. Is it because the source link is $g(x) = frac{d}{dx} f(x)$ is using scalar notation? He intends this to be the gradient of a vector
$endgroup$
– user3180
Nov 28 '18 at 8:01
$begingroup$
I don't know why you're talking about scalars, the Hessian H(x) is a vector and so is $Delta x$. Is it because the source link is $g(x) = frac{d}{dx} f(x)$ is using scalar notation? He intends this to be the gradient of a vector
$endgroup$
– user3180
Nov 28 '18 at 8:01
$begingroup$
user3180: to know if there is an error and if so if it is on lhs or rhs (hessian is a second order tensor not a vector, according to some definitions, or maybe even third order tensor)
$endgroup$
– Emil
Nov 28 '18 at 13:26
$begingroup$
user3180: to know if there is an error and if so if it is on lhs or rhs (hessian is a second order tensor not a vector, according to some definitions, or maybe even third order tensor)
$endgroup$
– Emil
Nov 28 '18 at 13:26
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
For a scalar function f,
$f(x+dx) approx f(x)+ frac{d f(x)}{dx}* dx$
Vector case [$x$ is (n,1) dimensional vector]:
$g(x + Delta x) approx g(x) + g(Delta x)$
$g(x) = nabla f(x) = begin{bmatrix}
g_1{(x)} \
g_2{(x)} \
vdots \
g_n{(x)}
end{bmatrix}$ $mid g_i{(x)} = frac{partial f(x)}{dx_i}$
$H(x) = nabla g(x) = begin{bmatrix}
nabla g_1{(x)}^T \
nabla g_2{(x)}^T \
vdots \
nabla g_n{(x)}^T
end{bmatrix}$
$g(x) +g(Delta x) = g(x) + Delta g(x)approx g(x)+ begin{bmatrix}
nabla g_1{(x)}^T Delta x \
nabla g_2{(x)}^T Delta x\
vdots \
nabla g_n{(x)}^T Delta x
end{bmatrix}$
Translation:
$g(Delta x)$ represents g with a small change in x as input, which is equivalent to $Delta g(x)$ [the infinitesimal change in g(x)].
If you look at the last column vector, you will see that each row $nabla g_i(x)^T Delta x$ represents the (rate of change of $g_i(x)$/rate of change of x) dotted with (infinitesimal change in x) := infinitesimal change in $g_i(x)$. In aggregate, the column vector represents the infinitesimal change in g(x) := $Delta g(x)$.
See also: http://www.friesian.com/calculus.htm
answered Nov 28 '18 at 9:48
user3180user3180
1185
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For a scalar function f,
$f(x+dx) approx f(x)+ frac{d f(x)}{dx}* dx$
Vector case [$x$ is (n,1) dimensional vector]:
$g(x + Delta x) approx g(x) + g(Delta x)$
$g(x) = nabla f(x) = begin{bmatrix}
g_1{(x)} \
g_2{(x)} \
vdots \
g_n{(x)}
end{bmatrix}$ $mid g_i{(x)} = frac{partial f(x)}{dx_i}$
$H(x) = nabla g(x) = begin{bmatrix}
nabla g_1{(x)}^T \
nabla g_2{(x)}^T \
vdots \
nabla g_n{(x)}^T
end{bmatrix}$
$g(x) +g(Delta x) = g(x) + Delta g(x)approx g(x)+ begin{bmatrix}
nabla g_1{(x)}^T Delta x \
nabla g_2{(x)}^T Delta x\
vdots \
nabla g_n{(x)}^T Delta x
end{bmatrix}$
Translation:
$g(Delta x)$ represents g with a small change in x as input, which is equivalent to $Delta g(x)$ [the infinitesimal change in g(x)].
If you look at the last column vector, you will see that each row $nabla g_i(x)^T Delta x$ represents the (rate of change of $g_i(x)$/rate of change of x) dotted with (infinitesimal change in x) := infinitesimal change in $g_i(x)$. In aggregate, the column vector represents the infinitesimal change in g(x) := $Delta g(x)$.
See also: http://www.friesian.com/calculus.htm
answered Nov 28 '18 at 9:48
user3180user3180
1185
$endgroup$
add a comment |
$begingroup$
For a scalar function f,
$f(x+dx) approx f(x)+ frac{d f(x)}{dx}* dx$
Vector case [$x$ is (n,1) dimensional vector]:
$g(x + Delta x) approx g(x) + g(Delta x)$
$g(x) = nabla f(x) = begin{bmatrix}
g_1{(x)} \
g_2{(x)} \
vdots \
g_n{(x)}
end{bmatrix}$ $mid g_i{(x)} = frac{partial f(x)}{dx_i}$
$H(x) = nabla g(x) = begin{bmatrix}
nabla g_1{(x)}^T \
nabla g_2{(x)}^T \
vdots \
nabla g_n{(x)}^T
end{bmatrix}$
$g(x) +g(Delta x) = g(x) + Delta g(x)approx g(x)+ begin{bmatrix}
nabla g_1{(x)}^T Delta x \
nabla g_2{(x)}^T Delta x\
vdots \
nabla g_n{(x)}^T Delta x
end{bmatrix}$
Translation:
$g(Delta x)$ represents g with a small change in x as input, which is equivalent to $Delta g(x)$ [the infinitesimal change in g(x)].
If you look at the last column vector, you will see that each row $nabla g_i(x)^T Delta x$ represents the (rate of change of $g_i(x)$/rate of change of x) dotted with (infinitesimal change in x) := infinitesimal change in $g_i(x)$. In aggregate, the column vector represents the infinitesimal change in g(x) := $Delta g(x)$.
See also: http://www.friesian.com/calculus.htm
answered Nov 28 '18 at 9:48
user3180user3180
1185
$endgroup$
add a comment |
$begingroup$
For a scalar function f,
$f(x+dx) approx f(x)+ frac{d f(x)}{dx}* dx$
Vector case [$x$ is (n,1) dimensional vector]:
$g(x + Delta x) approx g(x) + g(Delta x)$
$g(x) = nabla f(x) = begin{bmatrix}
g_1{(x)} \
g_2{(x)} \
vdots \
g_n{(x)}
end{bmatrix}$ $mid g_i{(x)} = frac{partial f(x)}{dx_i}$
$H(x) = nabla g(x) = begin{bmatrix}
nabla g_1{(x)}^T \
nabla g_2{(x)}^T \
vdots \
nabla g_n{(x)}^T
end{bmatrix}$
$g(x) +g(Delta x) = g(x) + Delta g(x)approx g(x)+ begin{bmatrix}
nabla g_1{(x)}^T Delta x \
nabla g_2{(x)}^T Delta x\
vdots \
nabla g_n{(x)}^T Delta x
end{bmatrix}$
Translation:
$g(Delta x)$ represents g with a small change in x as input, which is equivalent to $Delta g(x)$ [the infinitesimal change in g(x)].
If you look at the last column vector, you will see that each row $nabla g_i(x)^T Delta x$ represents the (rate of change of $g_i(x)$/rate of change of x) dotted with (infinitesimal change in x) := infinitesimal change in $g_i(x)$. In aggregate, the column vector represents the infinitesimal change in g(x) := $Delta g(x)$.
See also: http://www.friesian.com/calculus.htm
answered Nov 28 '18 at 9:48
user3180user3180
1185
$endgroup$
For a scalar function f,
$f(x+dx) approx f(x)+ frac{d f(x)}{dx}* dx$
Vector case [$x$ is (n,1) dimensional vector]:
$g(x + Delta x) approx g(x) + g(Delta x)$
$g(x) = nabla f(x) = begin{bmatrix}
g_1{(x)} \
g_2{(x)} \
vdots \
g_n{(x)}
end{bmatrix}$ $mid g_i{(x)} = frac{partial f(x)}{dx_i}$
$H(x) = nabla g(x) = begin{bmatrix}
nabla g_1{(x)}^T \
nabla g_2{(x)}^T \
vdots \
nabla g_n{(x)}^T
end{bmatrix}$
$g(x) +g(Delta x) = g(x) + Delta g(x)approx g(x)+ begin{bmatrix}
nabla g_1{(x)}^T Delta x \
nabla g_2{(x)}^T Delta x\
vdots \
nabla g_n{(x)}^T Delta x
end{bmatrix}$
Translation:
$g(Delta x)$ represents g with a small change in x as input, which is equivalent to $Delta g(x)$ [the infinitesimal change in g(x)].
If you look at the last column vector, you will see that each row $nabla g_i(x)^T Delta x$ represents the (rate of change of $g_i(x)$/rate of change of x) dotted with (infinitesimal change in x) := infinitesimal change in $g_i(x)$. In aggregate, the column vector represents the infinitesimal change in g(x) := $Delta g(x)$.
See also: http://www.friesian.com/calculus.htm
answered Nov 28 '18 at 9:48
user3180user3180
1185
edited Nov 28 '18 at 21:42
answered Nov 28 '18 at 9:48
user3180user3180
1185
answered Nov 28 '18 at 9:48
user3180user3180
1185
answered Nov 28 '18 at 9:48
user3180user3180
1185
1185
add a comment |
add a comment |
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$begingroup$
Is it like the third term in the taylor expansion? en.m.wikipedia.org/wiki/…
$endgroup$
– Emil
Nov 28 '18 at 7:14
$begingroup$
Is the lhs supposed to be the gradient taken at a point or is it a scalar? The hessian is a second order tensor, so it should eat two dx to become a scalar, not one, I believe.
$endgroup$
– Emil
Nov 28 '18 at 7:22
$begingroup$
$Delta x$ is a vector I believe
$endgroup$
– user3180
Nov 28 '18 at 7:58
$begingroup$
I don't know why you're talking about scalars, the Hessian H(x) is a vector and so is $Delta x$. Is it because the source link is $g(x) = frac{d}{dx} f(x)$ is using scalar notation? He intends this to be the gradient of a vector
$endgroup$
– user3180
Nov 28 '18 at 8:01
$begingroup$
user3180: to know if there is an error and if so if it is on lhs or rhs (hessian is a second order tensor not a vector, according to some definitions, or maybe even third order tensor)
$endgroup$
– Emil
Nov 28 '18 at 13:26