Convergence to a fixed point
$begingroup$
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
asked 5 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.
$endgroup$
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
asked 5 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
asked 5 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.
$endgroup$
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
analysis convergence numerical-methods fixed-point-theorems fixedpoints
asked 5 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 5 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.
edited 4 hours ago
Grace
asked 5 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 5 hours ago
Grace Grace
164
asked 5 hours ago
Grace Grace
164
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.
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.
Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
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 |
$begingroup$
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
$begingroup$
Another related question.
$endgroup$
– rtybase
4 hours ago
$begingroup$
Another related question.
$endgroup$
– rtybase
4 hours ago
1
1
$begingroup$
Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
$endgroup$
– rtybase
4 hours ago
$begingroup$
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
$begingroup$
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 Answers
3
active
oldest
votes
$begingroup$
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.
answered 4 hours ago
Robert ShoreRobert Shore
2,044116
$endgroup$
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
$$ x_n = f(x_{n-1})$$
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 |
$begingroup$
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.
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.
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%2f3138183%2fconvergence-to-a-fixed-point%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
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.
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.
answered 4 hours ago
Robert ShoreRobert Shore
2,044116
$endgroup$
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.
answered 4 hours ago
Robert ShoreRobert Shore
2,044116
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
2,044116
add a comment |
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
$$ x_n = f(x_{n-1})$$
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
$$ x_n = f(x_{n-1})$$
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
$$ x_n = f(x_{n-1})$$
answered 4 hours ago
Sean LeeSean Lee
483211
$endgroup$
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
$$ x_n = f(x_{n-1})$$
answered 4 hours ago
Sean LeeSean Lee
483211
answered 4 hours ago
Sean LeeSean Lee
483211
answered 4 hours ago
Sean LeeSean Lee
483211
answered 4 hours ago
Sean LeeSean Lee
483211
483211
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@RobertShore Yes, you are right, thanks!
$endgroup$
– Sean Lee
4 hours ago
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.
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.
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.
answered 3 hours ago
Kavi Rama MurthyKavi Rama Murthy
65k42766
$endgroup$
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.
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
answered 3 hours ago
Kavi Rama MurthyKavi Rama Murthy
65k42766
65k42766
add a comment |
add a comment |
Grace is a new contributor. Be nice, and check out our Code of Conduct.
Grace is a new contributor. Be nice, and check out our Code of Conduct.
Grace is a new contributor. Be nice, and check out our Code of Conduct.
Grace is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3138183%2fconvergence-to-a-fixed-point%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
$begingroup$
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