Help with a quadratic form
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I am trying to understand the following (it relates to a proof that $f(x) = phi(g(x))$ is convex if $g(x)$ is convex and $phi$ is increasing). After some manipulations I have:
$frac{partial^2 f}{partial x_jpartial x_i} = phi''(g(x))frac{partial g}{partial x_j} frac{partial g}{partial x_i} + phi'(g(x)) frac{partial^2 g}{partial x_jpartial x_i} $
Which (this is where I get lost) implies that:
$h^THf(x)h = phi''(g(x)) Sigma_{i,j}frac{partial g}{partial x_j} frac{partial g}{partial x_i}h_ih_j + phi'(g(x))h^TH(g(x))h$
$= phi''(g(x)) (Sigma_{i}frac{partial g}{partial x_i}h_i)^2 + phi'(g(x))h^TH(g(x))h$
I know that typically a quadratic form would be something like:
$x^TAx = Sigma_i Sigma_j a_{ij} x_ix_j$
I have trouble reconciling the typical form with what we have above, and have a few questions:
How do I reconcile the typical form and the form above? The $x$'s correspond to the $h$'s, but I cannot figure out the how the two partial terms end up there, when there is only a single constant term in the standard form.
It looks like there are two quadratic forms here (one for $phi''$ and another for $phi'$). Is that correct?
Why does the first form looks different to the second (written in summation notation rather than the Hessian)? Are they not equivalent?
Similarly, in the last equality, how can we get rid of the partials with respect to $j$ and keep the squared terms? I found this very confusing.
I would be grateful for any help. Thanks for your time.
calculus
asked Dec 4 at 1:07
Christian
227
add a comment |
up vote
0
down vote
favorite
I am trying to understand the following (it relates to a proof that $f(x) = phi(g(x))$ is convex if $g(x)$ is convex and $phi$ is increasing). After some manipulations I have:
$frac{partial^2 f}{partial x_jpartial x_i} = phi''(g(x))frac{partial g}{partial x_j} frac{partial g}{partial x_i} + phi'(g(x)) frac{partial^2 g}{partial x_jpartial x_i} $
Which (this is where I get lost) implies that:
$h^THf(x)h = phi''(g(x)) Sigma_{i,j}frac{partial g}{partial x_j} frac{partial g}{partial x_i}h_ih_j + phi'(g(x))h^TH(g(x))h$
$= phi''(g(x)) (Sigma_{i}frac{partial g}{partial x_i}h_i)^2 + phi'(g(x))h^TH(g(x))h$
I know that typically a quadratic form would be something like:
$x^TAx = Sigma_i Sigma_j a_{ij} x_ix_j$
I have trouble reconciling the typical form with what we have above, and have a few questions:
How do I reconcile the typical form and the form above? The $x$'s correspond to the $h$'s, but I cannot figure out the how the two partial terms end up there, when there is only a single constant term in the standard form.
It looks like there are two quadratic forms here (one for $phi''$ and another for $phi'$). Is that correct?
Why does the first form looks different to the second (written in summation notation rather than the Hessian)? Are they not equivalent?
Similarly, in the last equality, how can we get rid of the partials with respect to $j$ and keep the squared terms? I found this very confusing.
I would be grateful for any help. Thanks for your time.
calculus
asked Dec 4 at 1:07
Christian
227
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to understand the following (it relates to a proof that $f(x) = phi(g(x))$ is convex if $g(x)$ is convex and $phi$ is increasing). After some manipulations I have:
$frac{partial^2 f}{partial x_jpartial x_i} = phi''(g(x))frac{partial g}{partial x_j} frac{partial g}{partial x_i} + phi'(g(x)) frac{partial^2 g}{partial x_jpartial x_i} $
Which (this is where I get lost) implies that:
$h^THf(x)h = phi''(g(x)) Sigma_{i,j}frac{partial g}{partial x_j} frac{partial g}{partial x_i}h_ih_j + phi'(g(x))h^TH(g(x))h$
$= phi''(g(x)) (Sigma_{i}frac{partial g}{partial x_i}h_i)^2 + phi'(g(x))h^TH(g(x))h$
I know that typically a quadratic form would be something like:
$x^TAx = Sigma_i Sigma_j a_{ij} x_ix_j$
I have trouble reconciling the typical form with what we have above, and have a few questions:
How do I reconcile the typical form and the form above? The $x$'s correspond to the $h$'s, but I cannot figure out the how the two partial terms end up there, when there is only a single constant term in the standard form.
It looks like there are two quadratic forms here (one for $phi''$ and another for $phi'$). Is that correct?
Why does the first form looks different to the second (written in summation notation rather than the Hessian)? Are they not equivalent?
Similarly, in the last equality, how can we get rid of the partials with respect to $j$ and keep the squared terms? I found this very confusing.
I would be grateful for any help. Thanks for your time.
calculus
asked Dec 4 at 1:07
Christian
227
I am trying to understand the following (it relates to a proof that $f(x) = phi(g(x))$ is convex if $g(x)$ is convex and $phi$ is increasing). After some manipulations I have:
$frac{partial^2 f}{partial x_jpartial x_i} = phi''(g(x))frac{partial g}{partial x_j} frac{partial g}{partial x_i} + phi'(g(x)) frac{partial^2 g}{partial x_jpartial x_i} $
Which (this is where I get lost) implies that:
$h^THf(x)h = phi''(g(x)) Sigma_{i,j}frac{partial g}{partial x_j} frac{partial g}{partial x_i}h_ih_j + phi'(g(x))h^TH(g(x))h$
$= phi''(g(x)) (Sigma_{i}frac{partial g}{partial x_i}h_i)^2 + phi'(g(x))h^TH(g(x))h$
I know that typically a quadratic form would be something like:
$x^TAx = Sigma_i Sigma_j a_{ij} x_ix_j$
I have trouble reconciling the typical form with what we have above, and have a few questions:
How do I reconcile the typical form and the form above? The $x$'s correspond to the $h$'s, but I cannot figure out the how the two partial terms end up there, when there is only a single constant term in the standard form.
It looks like there are two quadratic forms here (one for $phi''$ and another for $phi'$). Is that correct?
Why does the first form looks different to the second (written in summation notation rather than the Hessian)? Are they not equivalent?
Similarly, in the last equality, how can we get rid of the partials with respect to $j$ and keep the squared terms? I found this very confusing.
I would be grateful for any help. Thanks for your time.
calculus
calculus
asked Dec 4 at 1:07
Christian
227
asked Dec 4 at 1:07
Christian
227
asked Dec 4 at 1:07
Christian
227
asked Dec 4 at 1:07
Christian
227
asked Dec 4 at 1:07
Christian
227
227
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