Ttyjukl

How to kill a localhost:8080

Can the Shape Water Cantrip be used to manipulate blood?

Specific Chinese carabiner QA?

If nine coins are tossed, what is the probability that the number of heads is even?

Caulking a corner instead of taping with joint compound?

Is there a way to find out the age of climbing ropes?

Is there a frame of reference in which I was born before I was conceived?

PTIJ: Should I stay away from my computer?

How do I deal with being envious of my own players?

Can we carry rice to Japan?

Would the melodic leap of the opening phrase of Mozart's K545 be considered dissonant?

Convergence to a fixed point

PTIJ: What’s wrong with eating meat and couscous?

Ahoy, Ye Traveler!

If there are any 3nion, 5nion, 7nion, 9nion, 10nion, etc.

Should we avoid writing fiction about historical events without extensive research?

Why would the IRS ask for birth certificates or even audit a small tax return?

Where is this quote about overcoming the impossible said in "Interstellar"?

Sometimes a banana is just a banana

How to fix my table, centering of columns

Giving a talk in my old university, how prominently should I tell students my salary?

How does signal strength relate to bandwidth?

When was drinking water recognized as crucial in marathon running?

“I had a flat in the centre of town, but I didn’t like living there, so …”

Convergence to a fixed point

Contraction mapping in the context of $f(x_n)=x_{n+1}$.Confused about fixed point method conditionShrinking Map and Fixed Point via Iteration MethodWhich negation of the definition of a null sequence is correct?Relation between two different definitions of quadratic convergenceFixed point, bounded derivativeProving Cauchy when given a sequenceBanach fixed point questionHelp me understand this proof of Implicit Function Theorem on Banach spacesConvergence of functions (in sense of distributions)Banach fixed-point theoremIf the odd function $f:mathbb Rtomathbb R$ letting $x>0$ is continuous at $x$, prove the function is continuous at $-x$.

3

$begingroup$

Let $f : [a,b] rightarrow [a,b]$ be a continuous function s.t. $f'(x)$ is defined on $(a,b)$ and $leftlvert f'(x)rightrvert leqq t$ where $0<t<1$. Prove that for any point $x_0$ in $[a,b]$ the sequence defined by $$ x_n=f(x_{n-1}), n>0$$
converges to one unique fixed point.

Attempt:
Frankly, I have struggled to make a real attempt due to the fact that I can't find notes relating to this.

Obviously, I am assuming that there exists $x$ in$ [a,b]$ s.t. $f(x)=x$ but how do I relate the sequence to this $x$?

I'm strictly not allowed to assume Banach's theorem in this question, nor the Cauchy sequence because they come up on the second part of the course. I rather have to PROVE this.

analysis convergence numerical-methods fixed-point-theorems fixedpoints

share|cite|improve this question

edited 3 hours ago

Grace

asked 4 hours ago

Grace Grace

164

New contributor

Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another related question.
  $endgroup$
  – rtybase
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
  $endgroup$
  – rtybase
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
  $endgroup$
  – dantopa
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Let $f : [a,b] rightarrow [a,b]$ be a continuous function s.t. $f'(x)$ is defined on $(a,b)$ and $leftlvert f'(x)rightrvert leqq t$ where $0<t<1$. Prove that for any point $x_0$ in $[a,b]$ the sequence defined by $$ x_n=f(x_{n-1}), n>0$$
converges to one unique fixed point.

Attempt:
Frankly, I have struggled to make a real attempt due to the fact that I can't find notes relating to this.

Obviously, I am assuming that there exists $x$ in$ [a,b]$ s.t. $f(x)=x$ but how do I relate the sequence to this $x$?

I'm strictly not allowed to assume Banach's theorem in this question, nor the Cauchy sequence because they come up on the second part of the course. I rather have to PROVE this.

analysis convergence numerical-methods fixed-point-theorems fixedpoints

share|cite|improve this question

edited 3 hours ago

Grace

asked 4 hours ago

Grace Grace

164

New contributor

Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another related question.
  $endgroup$
  – rtybase
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
  $endgroup$
  – rtybase
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
  $endgroup$
  – dantopa
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $f : [a,b] rightarrow [a,b]$ be a continuous function s.t. $f'(x)$ is defined on $(a,b)$ and $leftlvert f'(x)rightrvert leqq t$ where $0<t<1$. Prove that for any point $x_0$ in $[a,b]$ the sequence defined by $$ x_n=f(x_{n-1}), n>0$$
converges to one unique fixed point.

Attempt:
Frankly, I have struggled to make a real attempt due to the fact that I can't find notes relating to this.

Obviously, I am assuming that there exists $x$ in$ [a,b]$ s.t. $f(x)=x$ but how do I relate the sequence to this $x$?

I'm strictly not allowed to assume Banach's theorem in this question, nor the Cauchy sequence because they come up on the second part of the course. I rather have to PROVE this.

analysis convergence numerical-methods fixed-point-theorems fixedpoints

share|cite|improve this question

edited 3 hours ago

Grace

asked 4 hours ago

Grace Grace

164

New contributor

Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 3 hours ago

Grace

asked 4 hours ago

Grace Grace

164

New contributor

Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 3 hours ago

Grace

asked 4 hours ago

Grace Grace

164

New contributor

Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 4 hours ago

Grace Grace

164

New contributor

Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 4 hours ago

Grace Grace

164

3 Answers
3

active

oldest

votes

2

$begingroup$

The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

$endgroup$

add a comment | 

1

$begingroup$

This can be proved using the Banach Fixed Point Theorem.

Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
$$x_n = F(x_{n-1}) $$
will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.

Since in this case you know that
$$|f'(x)| leq t $$
This implies that

$$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$

for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).

This implies that

$$| f(x) - f(y)| < t |x-y| $$

which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence

$$ x_n = f(x_{n-1})$$

share|cite|improve this answer

answered 4 hours ago

Sean LeeSean Lee

483211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
  $endgroup$
  – Robert Shore
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertShore Yes, you are right, thanks!
  $endgroup$
  – Sean Lee
  4 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766

$endgroup$

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
});


}
});

Grace is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3138183%2fconvergence-to-a-fixed-point%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

$endgroup$

add a comment | 

2

$begingroup$

The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

$endgroup$

add a comment | 

$begingroup$

The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

1

$begingroup$

This can be proved using the Banach Fixed Point Theorem.

Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
$$x_n = F(x_{n-1}) $$
will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.

Since in this case you know that
$$|f'(x)| leq t $$
This implies that

$$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$

for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).

This implies that

$$| f(x) - f(y)| < t |x-y| $$

which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence

$$ x_n = f(x_{n-1})$$

share|cite|improve this answer

answered 4 hours ago

Sean LeeSean Lee

483211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
  $endgroup$
  – Robert Shore
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertShore Yes, you are right, thanks!
  $endgroup$
  – Sean Lee
  4 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

This can be proved using the Banach Fixed Point Theorem.

Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
$$x_n = F(x_{n-1}) $$
will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.

Since in this case you know that
$$|f'(x)| leq t $$
This implies that

$$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$

for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).

This implies that

$$| f(x) - f(y)| < t |x-y| $$

which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence

$$ x_n = f(x_{n-1})$$

share|cite|improve this answer

answered 4 hours ago

Sean LeeSean Lee

483211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
  $endgroup$
  – Robert Shore
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertShore Yes, you are right, thanks!
  $endgroup$
  – Sean Lee
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This can be proved using the Banach Fixed Point Theorem.

Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
$$x_n = F(x_{n-1}) $$
will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.

Since in this case you know that
$$|f'(x)| leq t $$
This implies that

$$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$

for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).

This implies that

$$| f(x) - f(y)| < t |x-y| $$

which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence

$$ x_n = f(x_{n-1})$$

share|cite|improve this answer

answered 4 hours ago

Sean LeeSean Lee

483211

$endgroup$

share|cite|improve this answer

answered 4 hours ago

Sean LeeSean Lee

483211

answered 4 hours ago

Sean LeeSean Lee

483211

answered 4 hours ago

Sean LeeSean Lee

483211

1

$begingroup$

To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766

$endgroup$

add a comment | 

1

$begingroup$

To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766

$endgroup$

add a comment | 

$begingroup$

To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766