If $xmapsto F(x)=int^x_af(s)ds$, then $F$ is $C^1$ and $F'(x)=f(x),;;forall;xin[a,b]$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












Assuming that beginalignf:[a,b]toBbbR^nendalign
is continuous, and
beginalignF:[a,b]toBbbR^nendalign
beginalignxmapsto F(x)=int^x_af(s)ds.endalign
I want to prove that $F$ is $C^1$ and $F'(x)=f(x),;;forall;xin[a,b].$



MY WORK



To prove differentiablity, it suffices to prove that beginalignVert F(x_0+h)- F(x_0)-hf(x_0)Vert leqVert h Vertepsilon(h)endalign



Since, $f:[a,b]toBbbR^n$, then it is uniformly continuous.



beginalignVert F(x_0+h)- F(x_0)-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_af(s)ds- int^x_0_af(s)ds-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_x_0f(s)ds-hf(x_0) Vertendalign
I'm stuck at this point, can anyone help me out?




calculus integration differential-equations multivariable-calculus derivatives




share|cite|improve this question
















  • 1




    Mean value theorem for integrals...
    – amsmath
    Aug 28 at 15:27






  • 2




    @amsmath Doesn't the Mean value theorem fail for vector-valued functions?
    – Sobi
    Aug 28 at 15:29







  • 1




    @Sobi Sure it does. I did not read carefully enough. My bad.
    – amsmath
    Aug 28 at 15:30










  • @amsmath Happened to me a million times! :)
    – Sobi
    Aug 28 at 15:31






  • 1




    @Sobi That happens because we wanna be the first to help, right? ;-)
    – amsmath
    Aug 28 at 18:02















up vote
1
down vote

favorite












Assuming that beginalignf:[a,b]toBbbR^nendalign
is continuous, and
beginalignF:[a,b]toBbbR^nendalign
beginalignxmapsto F(x)=int^x_af(s)ds.endalign
I want to prove that $F$ is $C^1$ and $F'(x)=f(x),;;forall;xin[a,b].$



MY WORK



To prove differentiablity, it suffices to prove that beginalignVert F(x_0+h)- F(x_0)-hf(x_0)Vert leqVert h Vertepsilon(h)endalign



Since, $f:[a,b]toBbbR^n$, then it is uniformly continuous.



beginalignVert F(x_0+h)- F(x_0)-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_af(s)ds- int^x_0_af(s)ds-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_x_0f(s)ds-hf(x_0) Vertendalign
I'm stuck at this point, can anyone help me out?




calculus integration differential-equations multivariable-calculus derivatives




share|cite|improve this question
















  • 1




    Mean value theorem for integrals...
    – amsmath
    Aug 28 at 15:27






  • 2




    @amsmath Doesn't the Mean value theorem fail for vector-valued functions?
    – Sobi
    Aug 28 at 15:29







  • 1




    @Sobi Sure it does. I did not read carefully enough. My bad.
    – amsmath
    Aug 28 at 15:30










  • @amsmath Happened to me a million times! :)
    – Sobi
    Aug 28 at 15:31






  • 1




    @Sobi That happens because we wanna be the first to help, right? ;-)
    – amsmath
    Aug 28 at 18:02













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Assuming that beginalignf:[a,b]toBbbR^nendalign
is continuous, and
beginalignF:[a,b]toBbbR^nendalign
beginalignxmapsto F(x)=int^x_af(s)ds.endalign
I want to prove that $F$ is $C^1$ and $F'(x)=f(x),;;forall;xin[a,b].$



MY WORK



To prove differentiablity, it suffices to prove that beginalignVert F(x_0+h)- F(x_0)-hf(x_0)Vert leqVert h Vertepsilon(h)endalign



Since, $f:[a,b]toBbbR^n$, then it is uniformly continuous.



beginalignVert F(x_0+h)- F(x_0)-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_af(s)ds- int^x_0_af(s)ds-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_x_0f(s)ds-hf(x_0) Vertendalign
I'm stuck at this point, can anyone help me out?




calculus integration differential-equations multivariable-calculus derivatives




share|cite|improve this question












Assuming that beginalignf:[a,b]toBbbR^nendalign
is continuous, and
beginalignF:[a,b]toBbbR^nendalign
beginalignxmapsto F(x)=int^x_af(s)ds.endalign
I want to prove that $F$ is $C^1$ and $F'(x)=f(x),;;forall;xin[a,b].$



MY WORK



To prove differentiablity, it suffices to prove that beginalignVert F(x_0+h)- F(x_0)-hf(x_0)Vert leqVert h Vertepsilon(h)endalign



Since, $f:[a,b]toBbbR^n$, then it is uniformly continuous.



beginalignVert F(x_0+h)- F(x_0)-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_af(s)ds- int^x_0_af(s)ds-hf(x_0) Vertendalign
beginalign=Vert int^x_0+h_x_0f(s)ds-hf(x_0) Vertendalign
I'm stuck at this point, can anyone help me out?





calculus integration differential-equations multivariable-calculus derivatives





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 28 at 15:20









Micheal

25710




25710







  • 1




    Mean value theorem for integrals...
    – amsmath
    Aug 28 at 15:27






  • 2




    @amsmath Doesn't the Mean value theorem fail for vector-valued functions?
    – Sobi
    Aug 28 at 15:29







  • 1




    @Sobi Sure it does. I did not read carefully enough. My bad.
    – amsmath
    Aug 28 at 15:30










  • @amsmath Happened to me a million times! :)
    – Sobi
    Aug 28 at 15:31






  • 1




    @Sobi That happens because we wanna be the first to help, right? ;-)
    – amsmath
    Aug 28 at 18:02













  • 1




    Mean value theorem for integrals...
    – amsmath
    Aug 28 at 15:27






  • 2




    @amsmath Doesn't the Mean value theorem fail for vector-valued functions?
    – Sobi
    Aug 28 at 15:29







  • 1




    @Sobi Sure it does. I did not read carefully enough. My bad.
    – amsmath
    Aug 28 at 15:30










  • @amsmath Happened to me a million times! :)
    – Sobi
    Aug 28 at 15:31






  • 1




    @Sobi That happens because we wanna be the first to help, right? ;-)
    – amsmath
    Aug 28 at 18:02








1




1




Mean value theorem for integrals...
– amsmath
Aug 28 at 15:27




Mean value theorem for integrals...
– amsmath
Aug 28 at 15:27




2




2




@amsmath Doesn't the Mean value theorem fail for vector-valued functions?
– Sobi
Aug 28 at 15:29





@amsmath Doesn't the Mean value theorem fail for vector-valued functions?
– Sobi
Aug 28 at 15:29





1




1




@Sobi Sure it does. I did not read carefully enough. My bad.
– amsmath
Aug 28 at 15:30




@Sobi Sure it does. I did not read carefully enough. My bad.
– amsmath
Aug 28 at 15:30












@amsmath Happened to me a million times! :)
– Sobi
Aug 28 at 15:31




@amsmath Happened to me a million times! :)
– Sobi
Aug 28 at 15:31




1




1




@Sobi That happens because we wanna be the first to help, right? ;-)
– amsmath
Aug 28 at 18:02





@Sobi That happens because we wanna be the first to help, right? ;-)
– amsmath
Aug 28 at 18:02











1 Answer
1






active

oldest

votes

















up vote
4
down vote



accepted










Use that the integral over a constant is the constant times the interval length, in reverse
beginalign
leftVert int^x_0+h_x_0f(s),ds-hf(x_0) rightVert
&=leftVert int^x_0+h_x_0(f(s)-f(x_0)),ds rightVert\
&le int^x_0+h_x_0Vert f(s)-f(x_0) Vert,ds
endalign






share|cite|improve this answer




















  • Wow, I love this!
    – Micheal
    Aug 28 at 15:28










  • Now, use uniform continuity of $f$ on $[a,b]$.
    – amsmath
    Aug 28 at 15:29










  • @amsmath: Done!
    – Micheal
    Aug 28 at 18:55










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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2897389%2fif-x-mapsto-fx-intx-afsds-then-f-is-c1-and-fx-fx-fo%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










Use that the integral over a constant is the constant times the interval length, in reverse
beginalign
leftVert int^x_0+h_x_0f(s),ds-hf(x_0) rightVert
&=leftVert int^x_0+h_x_0(f(s)-f(x_0)),ds rightVert\
&le int^x_0+h_x_0Vert f(s)-f(x_0) Vert,ds
endalign






share|cite|improve this answer




















  • Wow, I love this!
    – Micheal
    Aug 28 at 15:28










  • Now, use uniform continuity of $f$ on $[a,b]$.
    – amsmath
    Aug 28 at 15:29










  • @amsmath: Done!
    – Micheal
    Aug 28 at 18:55














up vote
4
down vote



accepted










Use that the integral over a constant is the constant times the interval length, in reverse
beginalign
leftVert int^x_0+h_x_0f(s),ds-hf(x_0) rightVert
&=leftVert int^x_0+h_x_0(f(s)-f(x_0)),ds rightVert\
&le int^x_0+h_x_0Vert f(s)-f(x_0) Vert,ds
endalign






share|cite|improve this answer




















  • Wow, I love this!
    – Micheal
    Aug 28 at 15:28










  • Now, use uniform continuity of $f$ on $[a,b]$.
    – amsmath
    Aug 28 at 15:29










  • @amsmath: Done!
    – Micheal
    Aug 28 at 18:55












up vote
4
down vote



accepted







up vote
4
down vote



accepted






Use that the integral over a constant is the constant times the interval length, in reverse
beginalign
leftVert int^x_0+h_x_0f(s),ds-hf(x_0) rightVert
&=leftVert int^x_0+h_x_0(f(s)-f(x_0)),ds rightVert\
&le int^x_0+h_x_0Vert f(s)-f(x_0) Vert,ds
endalign






share|cite|improve this answer












Use that the integral over a constant is the constant times the interval length, in reverse
beginalign
leftVert int^x_0+h_x_0f(s),ds-hf(x_0) rightVert
&=leftVert int^x_0+h_x_0(f(s)-f(x_0)),ds rightVert\
&le int^x_0+h_x_0Vert f(s)-f(x_0) Vert,ds
endalign







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 28 at 15:27









LutzL

50.1k31849




50.1k31849











  • Wow, I love this!
    – Micheal
    Aug 28 at 15:28










  • Now, use uniform continuity of $f$ on $[a,b]$.
    – amsmath
    Aug 28 at 15:29










  • @amsmath: Done!
    – Micheal
    Aug 28 at 18:55
















  • Wow, I love this!
    – Micheal
    Aug 28 at 15:28










  • Now, use uniform continuity of $f$ on $[a,b]$.
    – amsmath
    Aug 28 at 15:29










  • @amsmath: Done!
    – Micheal
    Aug 28 at 18:55















Wow, I love this!
– Micheal
Aug 28 at 15:28




Wow, I love this!
– Micheal
Aug 28 at 15:28












Now, use uniform continuity of $f$ on $[a,b]$.
– amsmath
Aug 28 at 15:29




Now, use uniform continuity of $f$ on $[a,b]$.
– amsmath
Aug 28 at 15:29












@amsmath: Done!
– Micheal
Aug 28 at 18:55




@amsmath: Done!
– Micheal
Aug 28 at 18:55

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2897389%2fif-x-mapsto-fx-intx-afsds-then-f-is-c1-and-fx-fx-fo%23new-answer', 'question_page');

);

Post as a guest
































































-

這個網誌中的熱門文章

How to combine Bézier curves to a surface?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
閱讀完整內容

Mutual Information Always Non-negative

圖片
Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
閱讀完整內容

Why am i infinitely getting the same tweet with the Twitter Search API?

圖片
up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
閱讀完整內容