Help with a quadratic form
$begingroup$
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 '18 at 1:07
ChristianChristian
367
$endgroup$
add a comment |
$begingroup$
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 '18 at 1:07
ChristianChristian
367
$endgroup$
add a comment |
$begingroup$
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 '18 at 1:07
ChristianChristian
367
$endgroup$
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 '18 at 1:07
ChristianChristian
367
asked Dec 4 '18 at 1:07
ChristianChristian
367
asked Dec 4 '18 at 1:07
ChristianChristian
367
asked Dec 4 '18 at 1:07
ChristianChristian
367
asked Dec 4 '18 at 1:07
ChristianChristian
367
367
add a comment |
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024971%2fhelp-with-a-quadratic-form%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024971%2fhelp-with-a-quadratic-form%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
