Convergence to a fixed point [duplicate]Contraction mapping in the context of $f(x_n)=x_{n+1}$.Confused about...
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Convergence to a fixed point [duplicate]
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 $-x$.
$begingroup$
This question already has an answer here:
Contraction mapping in the context of $f(x_n)=x_{n+1}$.
3 answers
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 17 hours ago
Grace Grace
255
New contributor
Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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marked as duplicate by rtybase, Eevee Trainer, mrtaurho, Vinyl_cape_jawa, José Carlos Santos 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Contraction mapping in the context of $f(x_n)=x_{n+1}$.
3 answers
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 17 hours ago
Grace Grace
255
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.
$endgroup$
marked as duplicate by rtybase, Eevee Trainer, mrtaurho, Vinyl_cape_jawa, José Carlos Santos 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Another related question.
$endgroup$
– rtybase
17 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.
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– dantopa
15 hours ago
add a comment |
$begingroup$
This question already has an answer here:
Contraction mapping in the context of $f(x_n)=x_{n+1}$.
3 answers
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 17 hours ago
Grace Grace
255
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.
$endgroup$
This question already has an answer here:
Contraction mapping in the context of $f(x_n)=x_{n+1}$.
3 answers
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.
This question already has an answer here:
Contraction mapping in the context of $f(x_n)=x_{n+1}$.
3 answers
analysis convergence numerical-methods fixed-point-theorems fixedpoints
analysis convergence numerical-methods fixed-point-theorems fixedpoints
asked 17 hours ago
Grace Grace
255
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 17 hours ago
Grace Grace
255
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 17 hours ago
Grace
asked 17 hours ago
Grace Grace
255
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 17 hours ago
Grace Grace
255
asked 17 hours ago
Grace Grace
255
255
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.
marked as duplicate by rtybase, Eevee Trainer, mrtaurho, Vinyl_cape_jawa, José Carlos Santos 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by rtybase, Eevee Trainer, mrtaurho, Vinyl_cape_jawa, José Carlos Santos 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Another related question.
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– rtybase
17 hours ago
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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.
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– dantopa
15 hours ago
add a comment |
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Another related question.
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– rtybase
17 hours ago
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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.
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– dantopa
15 hours ago
$begingroup$
Another related question.
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– rtybase
17 hours ago
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Another related question.
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– rtybase
17 hours ago
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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.
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– dantopa
15 hours ago
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15 hours ago
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3 Answers
3
active
oldest
votes
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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 16 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
$begingroup$
Thank you! This is very helpful :-)
$endgroup$
– Grace
7 hours ago
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 17 hours ago
Robert ShoreRobert Shore
2,079116
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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 17 hours ago
Sean LeeSean Lee
493211
$endgroup$
$begingroup$
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
17 hours ago
$begingroup$
@RobertShore Yes, you are right, thanks!
$endgroup$
– Sean Lee
17 hours ago
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 16 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
$begingroup$
Thank you! This is very helpful :-)
$endgroup$
– Grace
7 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 16 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
$begingroup$
Thank you! This is very helpful :-)
$endgroup$
– Grace
7 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 16 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
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 16 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
answered 16 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
answered 16 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
answered 16 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
65.4k42766
$begingroup$
Thank you! This is very helpful :-)
$endgroup$
– Grace
7 hours ago
add a comment |
$begingroup$
Thank you! This is very helpful :-)
$endgroup$
– Grace
7 hours ago
$begingroup$
Thank you! This is very helpful :-)
$endgroup$
– Grace
7 hours ago
$begingroup$
Thank you! This is very helpful :-)
$endgroup$
– Grace
7 hours ago
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 17 hours ago
Robert ShoreRobert Shore
2,079116
$endgroup$
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 17 hours ago
Robert ShoreRobert Shore
2,079116
$endgroup$
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 17 hours ago
Robert ShoreRobert Shore
2,079116
$endgroup$
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 17 hours ago
Robert ShoreRobert Shore
2,079116
edited 17 hours ago
answered 17 hours ago
Robert ShoreRobert Shore
2,079116
answered 17 hours ago
Robert ShoreRobert Shore
2,079116
answered 17 hours ago
Robert ShoreRobert Shore
2,079116
2,079116
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 17 hours ago
Sean LeeSean Lee
493211
$endgroup$
$begingroup$
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
17 hours ago
$begingroup$
@RobertShore Yes, you are right, thanks!
$endgroup$
– Sean Lee
17 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 17 hours ago
Sean LeeSean Lee
493211
$endgroup$
$begingroup$
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
17 hours ago
$begingroup$
@RobertShore Yes, you are right, thanks!
$endgroup$
– Sean Lee
17 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 17 hours ago
Sean LeeSean Lee
493211
$endgroup$
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 17 hours ago
Sean LeeSean Lee
493211
answered 17 hours ago
Sean LeeSean Lee
493211
answered 17 hours ago
Sean LeeSean Lee
493211
answered 17 hours ago
Sean LeeSean Lee
493211
493211
$begingroup$
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
17 hours ago
$begingroup$
@RobertShore Yes, you are right, thanks!
$endgroup$
– Sean Lee
17 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
17 hours ago
$begingroup$
@RobertShore Yes, you are right, thanks!
$endgroup$
– Sean Lee
17 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
17 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
17 hours ago
$begingroup$
@RobertShore Yes, you are right, thanks!
$endgroup$
– Sean Lee
17 hours ago
$begingroup$
@RobertShore Yes, you are right, thanks!
$endgroup$
– Sean Lee
17 hours ago
add a comment |
$begingroup$
Another related question.
$endgroup$
– rtybase
17 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
15 hours ago