Hessian-vector products












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 20:48







user3180

















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




















  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















0












$begingroup$

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






share|cite|improve this answer











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    1 Answer
    1






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    active

    oldest

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    0












    $begingroup$

    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






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      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






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        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






        share|cite|improve this answer











        $endgroup$



        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







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 28 '18 at 21:42

























        answered Nov 28 '18 at 9:48









        user3180user3180

        1185




        1185






























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