Intermediate value theorem with continuous functionIs function uniformly continuousThe product of a uniformly continuous function and a bounded continuous function is uniformly continuousExamples of unbounded continuous function $f:Qcap[0,1]rightarrow R$Find a continuous function on a closed interval with range an open intervalA continuous injective function and its inverseProving series of functions cannot sum to a continuous function.Measure of inverse image of a monotone function is continuous?Using Intermediate Value Theorem for continuous functionsFinding an $alpha$-Hölder continuous function $f_alpha$ at $0$, but not $beta$-Hölder.“Switching” of a continuous function
Why didn't Boeing produce its own regional jet?
How can I prove that a state of equilibrium is unstable?
Implication of namely
How to find if SQL server backup is encrypted with TDE without restoring the backup
What is an equivalently powerful replacement spell for the Yuan-Ti's Suggestion spell?
Does int main() need a declaration on C++?
Are British MPs missing the point, with these 'Indicative Votes'?
What are the G forces leaving Earth orbit?
How does a dynamic QR code work?
Can a virus destroy the BIOS of a modern computer?
Sums of two squares in arithmetic progressions
Send out email when Apex Queueable fails and test it
How to install cross-compiler on Ubuntu 18.04?
Unlock My Phone! February 2018
Notepad++ delete until colon for every line with replace all
Forgetting the musical notes while performing in concert
how do we prove that a sum of two periods is still a period?
How to compactly explain secondary and tertiary characters without resorting to stereotypes?
Is it possible to map the firing of neurons in the human brain so as to stimulate artificial memories in someone else?
How exploitable/balanced is this homebrew spell: Spell Permanency?
Processor speed limited at 0.4 Ghz
Bullying boss launched a smear campaign and made me unemployable
What is required to make GPS signals available indoors?
Could neural networks be considered metaheuristics?
Intermediate value theorem with continuous function
Is function uniformly continuousThe product of a uniformly continuous function and a bounded continuous function is uniformly continuousExamples of unbounded continuous function $f:Qcap[0,1]rightarrow R$Find a continuous function on a closed interval with range an open intervalA continuous injective function and its inverseProving series of functions cannot sum to a continuous function.Measure of inverse image of a monotone function is continuous?Using Intermediate Value Theorem for continuous functionsFinding an $alpha$-Hölder continuous function $f_alpha$ at $0$, but not $beta$-Hölder.“Switching” of a continuous function
$begingroup$
Let $f: [0,1] rightarrow mathbbR$ be continuous with $f(0)=f(1)$. If $h in (0,frac12)$ is not of the form $frac1n$, there does not necessarily exist $|x-y|=h$ satisfying $f(x)=f(y)$. Provide an example that illustrates this using $h=frac25$
I'm thinking I need to use some kind of modification to a sin function to get this to work. Not sure how to come up with an explicit formula, though. I could just draw a picture, but I want to find some explicit formula so I can show that it's true.
real-analysis continuity
asked Nov 2 '16 at 2:33
Determinant21Determinant21
616
$endgroup$
add a comment |
$begingroup$
Let $f: [0,1] rightarrow mathbbR$ be continuous with $f(0)=f(1)$. If $h in (0,frac12)$ is not of the form $frac1n$, there does not necessarily exist $|x-y|=h$ satisfying $f(x)=f(y)$. Provide an example that illustrates this using $h=frac25$
I'm thinking I need to use some kind of modification to a sin function to get this to work. Not sure how to come up with an explicit formula, though. I could just draw a picture, but I want to find some explicit formula so I can show that it's true.
real-analysis continuity
asked Nov 2 '16 at 2:33
Determinant21Determinant21
616
$endgroup$
add a comment |
$begingroup$
Let $f: [0,1] rightarrow mathbbR$ be continuous with $f(0)=f(1)$. If $h in (0,frac12)$ is not of the form $frac1n$, there does not necessarily exist $|x-y|=h$ satisfying $f(x)=f(y)$. Provide an example that illustrates this using $h=frac25$
I'm thinking I need to use some kind of modification to a sin function to get this to work. Not sure how to come up with an explicit formula, though. I could just draw a picture, but I want to find some explicit formula so I can show that it's true.
real-analysis continuity
asked Nov 2 '16 at 2:33
Determinant21Determinant21
616
$endgroup$
Let $f: [0,1] rightarrow mathbbR$ be continuous with $f(0)=f(1)$. If $h in (0,frac12)$ is not of the form $frac1n$, there does not necessarily exist $|x-y|=h$ satisfying $f(x)=f(y)$. Provide an example that illustrates this using $h=frac25$
I'm thinking I need to use some kind of modification to a sin function to get this to work. Not sure how to come up with an explicit formula, though. I could just draw a picture, but I want to find some explicit formula so I can show that it's true.
real-analysis continuity
real-analysis continuity
asked Nov 2 '16 at 2:33
Determinant21Determinant21
616
asked Nov 2 '16 at 2:33
Determinant21Determinant21
616
asked Nov 2 '16 at 2:33
Determinant21Determinant21
616
asked Nov 2 '16 at 2:33
Determinant21Determinant21
616
asked Nov 2 '16 at 2:33
Determinant21Determinant21
616
616
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you accept a piecewise function as an explicit formula, then the function in the figure below works.
answered Mar 28 at 12:43
Marwan MizuriMarwan Mizuri
18417
$endgroup$
add a comment |
$begingroup$
A modified $sin$ function could work, but I think it would be easier to reason using a piecewise linear function. Try defining a function like the one below. The $x$-intercepts are specifically chosen to be $0, frac16, frac13,frac12,frac23,frac56,1$. What has to be true about $|x-y|$ if $f(x)=f(y)$? (If you really want to use $sin$, then modify your $sin$ function so that it has $x$-intercepts at the same places as this function. That should also work.)
answered Nov 2 '16 at 2:58
kccukccu
10.8k11229
$endgroup$
$begingroup$
can you elaborate on how this fulfills the requirements? correct me if I'm wrong but since the distance between the peaks is 1/3, and the distance the intercepts x=0 and x=3/6 is 1/2, by continuity there are points x,y with |x-y|=h and f(x)=f(y)?
$endgroup$
– Alain
Jun 22 '17 at 5:51
$begingroup$
You are correct. As you can see I wrote this answer several months ago, and I don't remember what my reasoning was at the time. But now I think that any periodic function like this will not actually work.
$endgroup$
– kccu
Jun 24 '17 at 17:56
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%2f1995449%2fintermediate-value-theorem-with-continuous-function%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
$begingroup$
If you accept a piecewise function as an explicit formula, then the function in the figure below works.
answered Mar 28 at 12:43
Marwan MizuriMarwan Mizuri
18417
$endgroup$
add a comment |
$begingroup$
If you accept a piecewise function as an explicit formula, then the function in the figure below works.
answered Mar 28 at 12:43
Marwan MizuriMarwan Mizuri
18417
$endgroup$
add a comment |
$begingroup$
If you accept a piecewise function as an explicit formula, then the function in the figure below works.
answered Mar 28 at 12:43
Marwan MizuriMarwan Mizuri
18417
$endgroup$
If you accept a piecewise function as an explicit formula, then the function in the figure below works.
answered Mar 28 at 12:43
Marwan MizuriMarwan Mizuri
18417
answered Mar 28 at 12:43
Marwan MizuriMarwan Mizuri
18417
answered Mar 28 at 12:43
Marwan MizuriMarwan Mizuri
18417
answered Mar 28 at 12:43
Marwan MizuriMarwan Mizuri
18417
18417
add a comment |
add a comment |
$begingroup$
A modified $sin$ function could work, but I think it would be easier to reason using a piecewise linear function. Try defining a function like the one below. The $x$-intercepts are specifically chosen to be $0, frac16, frac13,frac12,frac23,frac56,1$. What has to be true about $|x-y|$ if $f(x)=f(y)$? (If you really want to use $sin$, then modify your $sin$ function so that it has $x$-intercepts at the same places as this function. That should also work.)
answered Nov 2 '16 at 2:58
kccukccu
10.8k11229
$endgroup$
$begingroup$
can you elaborate on how this fulfills the requirements? correct me if I'm wrong but since the distance between the peaks is 1/3, and the distance the intercepts x=0 and x=3/6 is 1/2, by continuity there are points x,y with |x-y|=h and f(x)=f(y)?
$endgroup$
– Alain
Jun 22 '17 at 5:51
$begingroup$
You are correct. As you can see I wrote this answer several months ago, and I don't remember what my reasoning was at the time. But now I think that any periodic function like this will not actually work.
$endgroup$
– kccu
Jun 24 '17 at 17:56
add a comment |
$begingroup$
A modified $sin$ function could work, but I think it would be easier to reason using a piecewise linear function. Try defining a function like the one below. The $x$-intercepts are specifically chosen to be $0, frac16, frac13,frac12,frac23,frac56,1$. What has to be true about $|x-y|$ if $f(x)=f(y)$? (If you really want to use $sin$, then modify your $sin$ function so that it has $x$-intercepts at the same places as this function. That should also work.)
answered Nov 2 '16 at 2:58
kccukccu
10.8k11229
$endgroup$
$begingroup$
can you elaborate on how this fulfills the requirements? correct me if I'm wrong but since the distance between the peaks is 1/3, and the distance the intercepts x=0 and x=3/6 is 1/2, by continuity there are points x,y with |x-y|=h and f(x)=f(y)?
$endgroup$
– Alain
Jun 22 '17 at 5:51
$begingroup$
You are correct. As you can see I wrote this answer several months ago, and I don't remember what my reasoning was at the time. But now I think that any periodic function like this will not actually work.
$endgroup$
– kccu
Jun 24 '17 at 17:56
add a comment |
$begingroup$
A modified $sin$ function could work, but I think it would be easier to reason using a piecewise linear function. Try defining a function like the one below. The $x$-intercepts are specifically chosen to be $0, frac16, frac13,frac12,frac23,frac56,1$. What has to be true about $|x-y|$ if $f(x)=f(y)$? (If you really want to use $sin$, then modify your $sin$ function so that it has $x$-intercepts at the same places as this function. That should also work.)
answered Nov 2 '16 at 2:58
kccukccu
10.8k11229
$endgroup$
A modified $sin$ function could work, but I think it would be easier to reason using a piecewise linear function. Try defining a function like the one below. The $x$-intercepts are specifically chosen to be $0, frac16, frac13,frac12,frac23,frac56,1$. What has to be true about $|x-y|$ if $f(x)=f(y)$? (If you really want to use $sin$, then modify your $sin$ function so that it has $x$-intercepts at the same places as this function. That should also work.)
answered Nov 2 '16 at 2:58
kccukccu
10.8k11229
answered Nov 2 '16 at 2:58
kccukccu
10.8k11229
answered Nov 2 '16 at 2:58
kccukccu
10.8k11229
answered Nov 2 '16 at 2:58
kccukccu
10.8k11229
10.8k11229
$begingroup$
can you elaborate on how this fulfills the requirements? correct me if I'm wrong but since the distance between the peaks is 1/3, and the distance the intercepts x=0 and x=3/6 is 1/2, by continuity there are points x,y with |x-y|=h and f(x)=f(y)?
$endgroup$
– Alain
Jun 22 '17 at 5:51
$begingroup$
You are correct. As you can see I wrote this answer several months ago, and I don't remember what my reasoning was at the time. But now I think that any periodic function like this will not actually work.
$endgroup$
– kccu
Jun 24 '17 at 17:56
add a comment |
$begingroup$
can you elaborate on how this fulfills the requirements? correct me if I'm wrong but since the distance between the peaks is 1/3, and the distance the intercepts x=0 and x=3/6 is 1/2, by continuity there are points x,y with |x-y|=h and f(x)=f(y)?
$endgroup$
– Alain
Jun 22 '17 at 5:51
$begingroup$
You are correct. As you can see I wrote this answer several months ago, and I don't remember what my reasoning was at the time. But now I think that any periodic function like this will not actually work.
$endgroup$
– kccu
Jun 24 '17 at 17:56
$begingroup$
can you elaborate on how this fulfills the requirements? correct me if I'm wrong but since the distance between the peaks is 1/3, and the distance the intercepts x=0 and x=3/6 is 1/2, by continuity there are points x,y with |x-y|=h and f(x)=f(y)?
$endgroup$
– Alain
Jun 22 '17 at 5:51
$begingroup$
can you elaborate on how this fulfills the requirements? correct me if I'm wrong but since the distance between the peaks is 1/3, and the distance the intercepts x=0 and x=3/6 is 1/2, by continuity there are points x,y with |x-y|=h and f(x)=f(y)?
$endgroup$
– Alain
Jun 22 '17 at 5:51
$begingroup$
You are correct. As you can see I wrote this answer several months ago, and I don't remember what my reasoning was at the time. But now I think that any periodic function like this will not actually work.
$endgroup$
– kccu
Jun 24 '17 at 17:56
$begingroup$
You are correct. As you can see I wrote this answer several months ago, and I don't remember what my reasoning was at the time. But now I think that any periodic function like this will not actually work.
$endgroup$
– kccu
Jun 24 '17 at 17:56
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%2f1995449%2fintermediate-value-theorem-with-continuous-function%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
