Graphical interpretation of mean value theorem
$begingroup$
I'm struggling understanding how this theorem works:
Let $fin C^1(Bsubseteqmathbb{R}^n;mathbb{R}^m)$. Let $x_0,xinmathring{B}$ such that the segment $Sinmathring{B}$ of extremes $x,x_0$ . Then
$||f(x)-f(x_0)||leqsup_{phiin S}||Df(phi)||cdot||x-x_0||$
What does this mean graphically? What I suppose is $frac{||f(x)-f(x_0)||}{||x-x_0||}$ is the secant line of extremes $x,x_0$, the slope of this line is $leq$ of a tangent line calculated in $phiin S$ but I don't understand why we use $sup$ (shouldn't $sup_{phiin S}$ be the extreme of the segment S which is $x$ or $x_0$?)
real-analysis
asked Dec 24 '18 at 15:35
ArchimedessArchimedess
256
$endgroup$
add a comment |
$begingroup$
I'm struggling understanding how this theorem works:
Let $fin C^1(Bsubseteqmathbb{R}^n;mathbb{R}^m)$. Let $x_0,xinmathring{B}$ such that the segment $Sinmathring{B}$ of extremes $x,x_0$ . Then
$||f(x)-f(x_0)||leqsup_{phiin S}||Df(phi)||cdot||x-x_0||$
What does this mean graphically? What I suppose is $frac{||f(x)-f(x_0)||}{||x-x_0||}$ is the secant line of extremes $x,x_0$, the slope of this line is $leq$ of a tangent line calculated in $phiin S$ but I don't understand why we use $sup$ (shouldn't $sup_{phiin S}$ be the extreme of the segment S which is $x$ or $x_0$?)
real-analysis
asked Dec 24 '18 at 15:35
ArchimedessArchimedess
256
$endgroup$
add a comment |
$begingroup$
I'm struggling understanding how this theorem works:
Let $fin C^1(Bsubseteqmathbb{R}^n;mathbb{R}^m)$. Let $x_0,xinmathring{B}$ such that the segment $Sinmathring{B}$ of extremes $x,x_0$ . Then
$||f(x)-f(x_0)||leqsup_{phiin S}||Df(phi)||cdot||x-x_0||$
What does this mean graphically? What I suppose is $frac{||f(x)-f(x_0)||}{||x-x_0||}$ is the secant line of extremes $x,x_0$, the slope of this line is $leq$ of a tangent line calculated in $phiin S$ but I don't understand why we use $sup$ (shouldn't $sup_{phiin S}$ be the extreme of the segment S which is $x$ or $x_0$?)
real-analysis
asked Dec 24 '18 at 15:35
ArchimedessArchimedess
256
$endgroup$
I'm struggling understanding how this theorem works:
Let $fin C^1(Bsubseteqmathbb{R}^n;mathbb{R}^m)$. Let $x_0,xinmathring{B}$ such that the segment $Sinmathring{B}$ of extremes $x,x_0$ . Then
$||f(x)-f(x_0)||leqsup_{phiin S}||Df(phi)||cdot||x-x_0||$
What does this mean graphically? What I suppose is $frac{||f(x)-f(x_0)||}{||x-x_0||}$ is the secant line of extremes $x,x_0$, the slope of this line is $leq$ of a tangent line calculated in $phiin S$ but I don't understand why we use $sup$ (shouldn't $sup_{phiin S}$ be the extreme of the segment S which is $x$ or $x_0$?)
real-analysis
real-analysis
asked Dec 24 '18 at 15:35
ArchimedessArchimedess
256
asked Dec 24 '18 at 15:35
ArchimedessArchimedess
256
asked Dec 24 '18 at 15:35
ArchimedessArchimedess
256
asked Dec 24 '18 at 15:35
ArchimedessArchimedess
256
asked Dec 24 '18 at 15:35
ArchimedessArchimedess
256
256
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We don't take the supremum of $phiin S$, we take the supremum over $phiin S$ of $|Df(phi)|$, which denotes the size of the derivative at $phi$. Intuitively, if the size of the derivative were smaller than $frac{|f(x)-f(x_0)|}{x-x_0}$ everywhere, then there would be no way to go from $f(x)$ to $f(x_0)$ over a segment of length $|x-x_0|$.
answered Dec 24 '18 at 15:45
SmileyCraftSmileyCraft
3,776519
$endgroup$
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%2f3051363%2fgraphical-interpretation-of-mean-value-theorem%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
$begingroup$
We don't take the supremum of $phiin S$, we take the supremum over $phiin S$ of $|Df(phi)|$, which denotes the size of the derivative at $phi$. Intuitively, if the size of the derivative were smaller than $frac{|f(x)-f(x_0)|}{x-x_0}$ everywhere, then there would be no way to go from $f(x)$ to $f(x_0)$ over a segment of length $|x-x_0|$.
answered Dec 24 '18 at 15:45
SmileyCraftSmileyCraft
3,776519
$endgroup$
add a comment |
$begingroup$
We don't take the supremum of $phiin S$, we take the supremum over $phiin S$ of $|Df(phi)|$, which denotes the size of the derivative at $phi$. Intuitively, if the size of the derivative were smaller than $frac{|f(x)-f(x_0)|}{x-x_0}$ everywhere, then there would be no way to go from $f(x)$ to $f(x_0)$ over a segment of length $|x-x_0|$.
answered Dec 24 '18 at 15:45
SmileyCraftSmileyCraft
3,776519
$endgroup$
add a comment |
$begingroup$
We don't take the supremum of $phiin S$, we take the supremum over $phiin S$ of $|Df(phi)|$, which denotes the size of the derivative at $phi$. Intuitively, if the size of the derivative were smaller than $frac{|f(x)-f(x_0)|}{x-x_0}$ everywhere, then there would be no way to go from $f(x)$ to $f(x_0)$ over a segment of length $|x-x_0|$.
answered Dec 24 '18 at 15:45
SmileyCraftSmileyCraft
3,776519
$endgroup$
We don't take the supremum of $phiin S$, we take the supremum over $phiin S$ of $|Df(phi)|$, which denotes the size of the derivative at $phi$. Intuitively, if the size of the derivative were smaller than $frac{|f(x)-f(x_0)|}{x-x_0}$ everywhere, then there would be no way to go from $f(x)$ to $f(x_0)$ over a segment of length $|x-x_0|$.
answered Dec 24 '18 at 15:45
SmileyCraftSmileyCraft
3,776519
answered Dec 24 '18 at 15:45
SmileyCraftSmileyCraft
3,776519
answered Dec 24 '18 at 15:45
SmileyCraftSmileyCraft
3,776519
answered Dec 24 '18 at 15:45
SmileyCraftSmileyCraft
3,776519
3,776519
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%2f3051363%2fgraphical-interpretation-of-mean-value-theorem%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
