Convergence to a fixed point [duplicate]Contraction mapping in the context of $f(x_n)=x_{n+1}$.Confused about...

Can a Mimic (container form) actually hold loot?

Is being socially reclusive okay for a graduate student?

Quitting employee has privileged access to critical information

Ultrafilters as a double dual

Why are special aircraft used for the carriers in the United States Navy?

I've given my players a lot of magic items. Is it reasonable for me to give them harder encounters?

Why is there an extra space when I type "ls" on the Desktop?

What is the purpose of a disclaimer like "this is not legal advice"?

Can you run a ground wire from stove directly to ground pole in the ground

3.5% Interest Student Loan or use all of my savings on Tuition?

Replacing tantalum capacitor with ceramic capacitor for Op Amps

Where is the fallacy here?

Are small insurances worth it

How can I be pwned if I'm not registered on the compromised site?

The (Easy) Road to Code

What's the best tool for cutting holes into duct work?

Why won't the strings command stop?

Sundering Titan and basic normal lands and snow lands

How do we objectively assess if a dialogue sounds unnatural or cringy?

What is the oldest European royal house?

Is there a math equivalent to the conditional ternary operator?

Should I use HTTPS on a domain that will only be used for redirection?

Was it really inappropriate to write a pull request for the company I interviewed with?

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



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 $-x$.













4












$begingroup$



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.










share|cite|improve this question









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 14 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
    yesterday










  • $begingroup$
    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
    yesterday


















4












$begingroup$



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.










share|cite|improve this question









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 14 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
    yesterday










  • $begingroup$
    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
    yesterday
















4












4








4





$begingroup$



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.










share|cite|improve this question









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






share|cite|improve this question









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









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




share|cite|improve this question








edited yesterday







Grace













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 yesterday









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 14 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 14 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
    yesterday










  • $begingroup$
    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
    yesterday




















  • $begingroup$
    Another related question.
    $endgroup$
    – rtybase
    yesterday










  • $begingroup$
    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
    yesterday


















$begingroup$
Another related question.
$endgroup$
– rtybase
yesterday




$begingroup$
Another related question.
$endgroup$
– rtybase
yesterday












$begingroup$
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
yesterday






$begingroup$
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
yesterday












3 Answers
3






active

oldest

votes


















3












$begingroup$

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









$endgroup$













  • $begingroup$
    Thank you! This is very helpful :-)
    $endgroup$
    – Grace
    19 hours ago



















3












$begingroup$

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











$endgroup$





















    2












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      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
      yesterday










    • $begingroup$
      @RobertShore Yes, you are right, thanks!
      $endgroup$
      – Sean Lee
      yesterday


















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      Thank you! This is very helpful :-)
      $endgroup$
      – Grace
      19 hours ago
















    3












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      Thank you! This is very helpful :-)
      $endgroup$
      – Grace
      19 hours ago














    3












    3








    3





    $begingroup$

    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









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered yesterday









    Kavi Rama MurthyKavi Rama Murthy

    65.4k42766




    65.4k42766












    • $begingroup$
      Thank you! This is very helpful :-)
      $endgroup$
      – Grace
      19 hours ago


















    • $begingroup$
      Thank you! This is very helpful :-)
      $endgroup$
      – Grace
      19 hours ago
















    $begingroup$
    Thank you! This is very helpful :-)
    $endgroup$
    – Grace
    19 hours ago




    $begingroup$
    Thank you! This is very helpful :-)
    $endgroup$
    – Grace
    19 hours ago











    3












    $begingroup$

    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











    $endgroup$


















      3












      $begingroup$

      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











      $endgroup$
















        3












        3








        3





        $begingroup$

        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











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered yesterday









        Robert ShoreRobert Shore

        2,119116




        2,119116























            2












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              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
              yesterday










            • $begingroup$
              @RobertShore Yes, you are right, thanks!
              $endgroup$
              – Sean Lee
              yesterday
















            2












            $begingroup$

            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









            $endgroup$













            • $begingroup$
              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
              yesterday










            • $begingroup$
              @RobertShore Yes, you are right, thanks!
              $endgroup$
              – Sean Lee
              yesterday














            2












            2








            2





            $begingroup$

            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









            $endgroup$



            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



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







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            Sean LeeSean Lee

            503211




            503211












            • $begingroup$
              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
              yesterday










            • $begingroup$
              @RobertShore Yes, you are right, thanks!
              $endgroup$
              – Sean Lee
              yesterday


















            • $begingroup$
              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
              yesterday










            • $begingroup$
              @RobertShore Yes, you are right, thanks!
              $endgroup$
              – Sean Lee
              yesterday
















            $begingroup$
            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
            yesterday




            $begingroup$
            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
            yesterday












            $begingroup$
            @RobertShore Yes, you are right, thanks!
            $endgroup$
            – Sean Lee
            yesterday




            $begingroup$
            @RobertShore Yes, you are right, thanks!
            $endgroup$
            – Sean Lee
            yesterday



            Popular posts from this blog

            VNC viewer RFB protocol error: bad desktop size 0x0I Cannot Type the Key 'd' (lowercase) in VNC Viewer...

            Couldn't open a raw socket. Error: Permission denied (13) (nmap)Is it possible to run networking commands...

            Why not use the yoke to control yaw, as well as pitch and roll? Announcing the arrival of...