Why is $nabla f({bf p}) =sum_i left( 2{bf p} - 2{bf p}_i right) $?
$begingroup$
If
$$
f({bf p}) = sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_iright)
$$
Why is the gradient of $f(p)$ equal to
$
nabla f({bf p}) = frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right)
=sum_i left( 2{bf p} - 2{bf p}_i right) $
calculus multivariable-calculus
asked Dec 5 '18 at 1:08
K.MK.M
693412
$endgroup$
add a comment |
$begingroup$
If
$$
f({bf p}) = sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_iright)
$$
Why is the gradient of $f(p)$ equal to
$
nabla f({bf p}) = frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right)
=sum_i left( 2{bf p} - 2{bf p}_i right) $
calculus multivariable-calculus
asked Dec 5 '18 at 1:08
K.MK.M
693412
$endgroup$
add a comment |
$begingroup$
If
$$
f({bf p}) = sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_iright)
$$
Why is the gradient of $f(p)$ equal to
$
nabla f({bf p}) = frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right)
=sum_i left( 2{bf p} - 2{bf p}_i right) $
calculus multivariable-calculus
asked Dec 5 '18 at 1:08
K.MK.M
693412
$endgroup$
If
$$
f({bf p}) = sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_iright)
$$
Why is the gradient of $f(p)$ equal to
$
nabla f({bf p}) = frac{partial}{partial {bf p}}sum_i left({bf p}cdot {bf p} - 2{bf p}cdot {bf p}_i + {bf p}_icdot {bf p}_i right)
=sum_i left( 2{bf p} - 2{bf p}_i right) $
calculus multivariable-calculus
calculus multivariable-calculus
asked Dec 5 '18 at 1:08
K.MK.M
693412
asked Dec 5 '18 at 1:08
K.MK.M
693412
asked Dec 5 '18 at 1:08
K.MK.M
693412
asked Dec 5 '18 at 1:08
K.MK.M
693412
asked Dec 5 '18 at 1:08
K.MK.M
693412
693412
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can do it in components, consider the first term for example
$$
{bf p}cdot {bf p} = p_1p_1 + p_2p_2 + cdots
$$
So the gradient of this term is
begin{eqnarray}
frac{partial {bf p}cdot {bf p}}{partial p_1} &=& 2p_1 \
frac{partial {bf p}cdot {bf p}}{partial p_2} &=& 2p_2 \
&vdots &
end{eqnarray}
or in general
$$
frac{partial {bf p}cdot {bf p}}{partial {bf p}} = 2{bf p}
$$
Similarly you can prove that
$$
frac{partial {bf p}cdot {bf p}_i}{partial {bf p}} = {bf p}_i
$$
answered Dec 5 '18 at 1:45
caveraccaverac
14.5k31130
$endgroup$
add a comment |
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%2f3026450%2fwhy-is-nabla-f-bf-p-sum-i-left-2-bf-p-2-bf-p-i-right%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can do it in components, consider the first term for example
$$
{bf p}cdot {bf p} = p_1p_1 + p_2p_2 + cdots
$$
So the gradient of this term is
begin{eqnarray}
frac{partial {bf p}cdot {bf p}}{partial p_1} &=& 2p_1 \
frac{partial {bf p}cdot {bf p}}{partial p_2} &=& 2p_2 \
&vdots &
end{eqnarray}
or in general
$$
frac{partial {bf p}cdot {bf p}}{partial {bf p}} = 2{bf p}
$$
Similarly you can prove that
$$
frac{partial {bf p}cdot {bf p}_i}{partial {bf p}} = {bf p}_i
$$
answered Dec 5 '18 at 1:45
caveraccaverac
14.5k31130
$endgroup$
add a comment |
$begingroup$
You can do it in components, consider the first term for example
$$
{bf p}cdot {bf p} = p_1p_1 + p_2p_2 + cdots
$$
So the gradient of this term is
begin{eqnarray}
frac{partial {bf p}cdot {bf p}}{partial p_1} &=& 2p_1 \
frac{partial {bf p}cdot {bf p}}{partial p_2} &=& 2p_2 \
&vdots &
end{eqnarray}
or in general
$$
frac{partial {bf p}cdot {bf p}}{partial {bf p}} = 2{bf p}
$$
Similarly you can prove that
$$
frac{partial {bf p}cdot {bf p}_i}{partial {bf p}} = {bf p}_i
$$
answered Dec 5 '18 at 1:45
caveraccaverac
14.5k31130
$endgroup$
add a comment |
$begingroup$
You can do it in components, consider the first term for example
$$
{bf p}cdot {bf p} = p_1p_1 + p_2p_2 + cdots
$$
So the gradient of this term is
begin{eqnarray}
frac{partial {bf p}cdot {bf p}}{partial p_1} &=& 2p_1 \
frac{partial {bf p}cdot {bf p}}{partial p_2} &=& 2p_2 \
&vdots &
end{eqnarray}
or in general
$$
frac{partial {bf p}cdot {bf p}}{partial {bf p}} = 2{bf p}
$$
Similarly you can prove that
$$
frac{partial {bf p}cdot {bf p}_i}{partial {bf p}} = {bf p}_i
$$
answered Dec 5 '18 at 1:45
caveraccaverac
14.5k31130
$endgroup$
You can do it in components, consider the first term for example
$$
{bf p}cdot {bf p} = p_1p_1 + p_2p_2 + cdots
$$
So the gradient of this term is
begin{eqnarray}
frac{partial {bf p}cdot {bf p}}{partial p_1} &=& 2p_1 \
frac{partial {bf p}cdot {bf p}}{partial p_2} &=& 2p_2 \
&vdots &
end{eqnarray}
or in general
$$
frac{partial {bf p}cdot {bf p}}{partial {bf p}} = 2{bf p}
$$
Similarly you can prove that
$$
frac{partial {bf p}cdot {bf p}_i}{partial {bf p}} = {bf p}_i
$$
answered Dec 5 '18 at 1:45
caveraccaverac
14.5k31130
edited Dec 5 '18 at 1:51
answered Dec 5 '18 at 1:45
caveraccaverac
14.5k31130
answered Dec 5 '18 at 1:45
caveraccaverac
14.5k31130
answered Dec 5 '18 at 1:45
caveraccaverac
14.5k31130
14.5k31130
add a comment |
add a comment |
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%2f3026450%2fwhy-is-nabla-f-bf-p-sum-i-left-2-bf-p-2-bf-p-i-right%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
