Is the gradient of the self-intersections of a curve zero? The 2019 Stack Overflow Developer Survey Results Are InMonotonic curvature and self intersections.Parallel translation along a self intersecting curveSelf adjoint total covariant derivativeStokes Theorem for Manifolds with Self-IntersectionsIntersections of two curves in $mathbbR^n$Self intersections of a smooth closed curve being deformedProving that strictly monotonic curvature implies no self intersections (more specifically, using the following inequalities)Does an immersed curve in general position has finite self-intersections?Can we describe Injective and non-Injective functions through intersections?Problem understanding the gradient of a field.
Unbreakable Formation vs. Cry of the Carnarium
Could JWST stay at L2 "forever"?
Does it makes sense to buy a new cycle to learn riding?
Is domain driven design an anti-SQL pattern?
What are the advantages and disadvantages of running one shots compared to campaigns?
Why can Shazam do this?
What is the motivation for a law requiring 2 parties to consent for recording a conversation
Can we apply L'Hospital's rule?
Can I write a for loop that iterates over both collections and arrays?
What is this 4-propeller plane?
Is flight data recorder erased after every flight?
Pristine Bit Checking
Inline version of a function returns different value then non-inline version
Does light intensity oscillate really fast since it is a wave?
Why is the design of haulage companies so “special”?
If you're not a professional, what motivates you to keep writing?
A poker game description that does not feel gimmicky
Inflated grade on resume at previous job, might former employer tell new employer?
Who is that cowgirl appearing during the Columbia Pictures intro?
Spanish for "widget"
Springs with some finite mass
Did USCIS resume its biometric service for UK visa
How was Skylab's orbit inclination chosen?
Does a dangling wire really electrocute me if I'm standing in water?
Is the gradient of the self-intersections of a curve zero?
The 2019 Stack Overflow Developer Survey Results Are InMonotonic curvature and self intersections.Parallel translation along a self intersecting curveSelf adjoint total covariant derivativeStokes Theorem for Manifolds with Self-IntersectionsIntersections of two curves in $mathbbR^n$Self intersections of a smooth closed curve being deformedProving that strictly monotonic curvature implies no self intersections (more specifically, using the following inequalities)Does an immersed curve in general position has finite self-intersections?Can we describe Injective and non-Injective functions through intersections?Problem understanding the gradient of a field.
$begingroup$
Suppose a curve with self-intersections can be described by $phi(x,y)=0$. Suppose the intersections are $T_i$, $i=1,2,...$ and the gradient $nabla phi$ at those intersections are well defined. Then is it true that $nablaphi(T_i)=0$ for all $i$? In other words, are the gradients at those intersections all zero?
real-analysis calculus geometry differential-geometry
edited 1 hour ago
Ernie060
2,940719
asked 2 hours ago
winstonwinston
537418
$endgroup$
add a comment |
$begingroup$
Suppose a curve with self-intersections can be described by $phi(x,y)=0$. Suppose the intersections are $T_i$, $i=1,2,...$ and the gradient $nabla phi$ at those intersections are well defined. Then is it true that $nablaphi(T_i)=0$ for all $i$? In other words, are the gradients at those intersections all zero?
real-analysis calculus geometry differential-geometry
edited 1 hour ago
Ernie060
2,940719
asked 2 hours ago
winstonwinston
537418
$endgroup$
add a comment |
$begingroup$
Suppose a curve with self-intersections can be described by $phi(x,y)=0$. Suppose the intersections are $T_i$, $i=1,2,...$ and the gradient $nabla phi$ at those intersections are well defined. Then is it true that $nablaphi(T_i)=0$ for all $i$? In other words, are the gradients at those intersections all zero?
real-analysis calculus geometry differential-geometry
edited 1 hour ago
Ernie060
2,940719
asked 2 hours ago
winstonwinston
537418
$endgroup$
Suppose a curve with self-intersections can be described by $phi(x,y)=0$. Suppose the intersections are $T_i$, $i=1,2,...$ and the gradient $nabla phi$ at those intersections are well defined. Then is it true that $nablaphi(T_i)=0$ for all $i$? In other words, are the gradients at those intersections all zero?
real-analysis calculus geometry differential-geometry
real-analysis calculus geometry differential-geometry
edited 1 hour ago
Ernie060
2,940719
asked 2 hours ago
winstonwinston
537418
edited 1 hour ago
Ernie060
2,940719
asked 2 hours ago
winstonwinston
537418
edited 1 hour ago
Ernie060
2,940719
edited 1 hour ago
Ernie060
2,940719
edited 1 hour ago
Ernie060
2,940719
2,940719
asked 2 hours ago
winstonwinston
537418
asked 2 hours ago
winstonwinston
537418
asked 2 hours ago
winstonwinston
537418
537418
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming $phi(x,y)$ is continuously differentiable in a neighbourhood of $T_i$, yes, because otherwise you could use the Implicit Function Theorem to get a unique curve in a neighourhood of $T_i$ satisfying $phi(x,y) = 0$.
answered 1 hour ago
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
$begingroup$
If we agree that $phi$ is continuously differentiable (so $nabla phi(x,y)$ is a continuous function of $x$ and $y$), then yes, this must be true.
The reason is that, if $nabla phi(x_0, y_0) neq 0$ for some $(x_0, y_0)$, then the implicit function theorem guarantees that (locally) we can write $y$ as a function of $x$ or $x$ as a function of $y$. However, at a self-intersection $T_i$, our curve fails the horizontal and vertical line tests, so we cannot express $x$ as a function of $y$ or $y$ as a function of $x$.
answered 1 hour ago
StrantsStrants
5,84921736
$endgroup$
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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%2f3181199%2fis-the-gradient-of-the-self-intersections-of-a-curve-zero%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming $phi(x,y)$ is continuously differentiable in a neighbourhood of $T_i$, yes, because otherwise you could use the Implicit Function Theorem to get a unique curve in a neighourhood of $T_i$ satisfying $phi(x,y) = 0$.
answered 1 hour ago
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
$begingroup$
Assuming $phi(x,y)$ is continuously differentiable in a neighbourhood of $T_i$, yes, because otherwise you could use the Implicit Function Theorem to get a unique curve in a neighourhood of $T_i$ satisfying $phi(x,y) = 0$.
answered 1 hour ago
Robert IsraelRobert Israel
331k23220475
$endgroup$
add a comment |
$begingroup$
Assuming $phi(x,y)$ is continuously differentiable in a neighbourhood of $T_i$, yes, because otherwise you could use the Implicit Function Theorem to get a unique curve in a neighourhood of $T_i$ satisfying $phi(x,y) = 0$.
answered 1 hour ago
Robert IsraelRobert Israel
331k23220475
$endgroup$
Assuming $phi(x,y)$ is continuously differentiable in a neighbourhood of $T_i$, yes, because otherwise you could use the Implicit Function Theorem to get a unique curve in a neighourhood of $T_i$ satisfying $phi(x,y) = 0$.
answered 1 hour ago
Robert IsraelRobert Israel
331k23220475
answered 1 hour ago
Robert IsraelRobert Israel
331k23220475
answered 1 hour ago
Robert IsraelRobert Israel
331k23220475
answered 1 hour ago
Robert IsraelRobert Israel
331k23220475
331k23220475
add a comment |
add a comment |
$begingroup$
If we agree that $phi$ is continuously differentiable (so $nabla phi(x,y)$ is a continuous function of $x$ and $y$), then yes, this must be true.
The reason is that, if $nabla phi(x_0, y_0) neq 0$ for some $(x_0, y_0)$, then the implicit function theorem guarantees that (locally) we can write $y$ as a function of $x$ or $x$ as a function of $y$. However, at a self-intersection $T_i$, our curve fails the horizontal and vertical line tests, so we cannot express $x$ as a function of $y$ or $y$ as a function of $x$.
answered 1 hour ago
StrantsStrants
5,84921736
$endgroup$
add a comment |
$begingroup$
If we agree that $phi$ is continuously differentiable (so $nabla phi(x,y)$ is a continuous function of $x$ and $y$), then yes, this must be true.
The reason is that, if $nabla phi(x_0, y_0) neq 0$ for some $(x_0, y_0)$, then the implicit function theorem guarantees that (locally) we can write $y$ as a function of $x$ or $x$ as a function of $y$. However, at a self-intersection $T_i$, our curve fails the horizontal and vertical line tests, so we cannot express $x$ as a function of $y$ or $y$ as a function of $x$.
answered 1 hour ago
StrantsStrants
5,84921736
$endgroup$
add a comment |
$begingroup$
If we agree that $phi$ is continuously differentiable (so $nabla phi(x,y)$ is a continuous function of $x$ and $y$), then yes, this must be true.
The reason is that, if $nabla phi(x_0, y_0) neq 0$ for some $(x_0, y_0)$, then the implicit function theorem guarantees that (locally) we can write $y$ as a function of $x$ or $x$ as a function of $y$. However, at a self-intersection $T_i$, our curve fails the horizontal and vertical line tests, so we cannot express $x$ as a function of $y$ or $y$ as a function of $x$.
answered 1 hour ago
StrantsStrants
5,84921736
$endgroup$
If we agree that $phi$ is continuously differentiable (so $nabla phi(x,y)$ is a continuous function of $x$ and $y$), then yes, this must be true.
The reason is that, if $nabla phi(x_0, y_0) neq 0$ for some $(x_0, y_0)$, then the implicit function theorem guarantees that (locally) we can write $y$ as a function of $x$ or $x$ as a function of $y$. However, at a self-intersection $T_i$, our curve fails the horizontal and vertical line tests, so we cannot express $x$ as a function of $y$ or $y$ as a function of $x$.
answered 1 hour ago
StrantsStrants
5,84921736
answered 1 hour ago
StrantsStrants
5,84921736
answered 1 hour ago
StrantsStrants
5,84921736
answered 1 hour ago
StrantsStrants
5,84921736
5,84921736
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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%2f3181199%2fis-the-gradient-of-the-self-intersections-of-a-curve-zero%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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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
