Convergence to a fixed point












4












$begingroup$


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







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  • $begingroup$
    Another related question.
    $endgroup$
    – rtybase
    6 hours ago






  • 1




    $begingroup$
    Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
    $endgroup$
    – rtybase
    6 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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    4 hours ago


















4












$begingroup$


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









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
    6 hours ago






  • 1




    $begingroup$
    Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
    $endgroup$
    – rtybase
    6 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
    4 hours ago
















4












4








4





$begingroup$


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









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






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.
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share|cite|improve this question




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edited 6 hours ago







Grace













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asked 7 hours ago









Grace Grace

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  • $begingroup$
    Another related question.
    $endgroup$
    – rtybase
    6 hours ago






  • 1




    $begingroup$
    Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
    $endgroup$
    – rtybase
    6 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
    4 hours ago




















  • $begingroup$
    Another related question.
    $endgroup$
    – rtybase
    6 hours ago






  • 1




    $begingroup$
    Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
    $endgroup$
    – rtybase
    6 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
    4 hours ago


















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




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




1




1




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




$begingroup$
Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
$endgroup$
– rtybase
6 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
4 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
4 hours ago












3 Answers
3






active

oldest

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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
      6 hours ago










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



















    2












    $begingroup$

    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









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      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 6 hours ago

























          answered 6 hours ago









          Robert ShoreRobert Shore

          2,054116




          2,054116























              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
                6 hours ago










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
















              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
                6 hours ago










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














              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 6 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
                6 hours ago










              • $begingroup$
                @RobertShore Yes, you are right, thanks!
                $endgroup$
                – Sean Lee
                6 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
                6 hours ago










              • $begingroup$
                @RobertShore Yes, you are right, thanks!
                $endgroup$
                – Sean Lee
                6 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
              6 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
              6 hours ago












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




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











              2












              $begingroup$

              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









              $endgroup$


















                2












                $begingroup$

                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









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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 5 hours ago









                  Kavi Rama MurthyKavi Rama Murthy

                  65.1k42766




                  65.1k42766






















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                      How to make a Squid Proxy server?

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                      1 I want to set up squid proxy server for monitoring user activity in a network and Internet Web URL filtering for user. I also want to configure the user login, user log, user access group, and block unwanted web sites in my network. Please help me and guide me the full process to do this. server proxy squid share | improve this question edited Jul 31 '12 at 14:45 Jorge Castro 37.1k 107 422 617 asked Jul 31 '12 at 5:16 Goivind Tiwari Goivind Tiwari
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                      Is this a new Fibonacci Identity?

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                      6 1 $begingroup$ I have found the following Fibonacci Identity (and proved it). If $F_n$ denotes the nth Fibonacci Number, we have the following identity begin{equation} F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1} end{equation} where $F_1 = F_2 = 1$ , $r leq n$ , $h leq g$ , and $n, g, k in mathbb{N}$ . It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few. I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see w
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                      19世紀

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                      この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2012年3月 ） 千年紀: 2千年紀 世紀: 18世紀 - 19世紀 - 20世紀 十年紀: 1800年代 1810年代 1820年代 1830年代 1840年代 1850年代 1860年代 1870年代 1880年代 1890年代 19世紀に君臨した大英帝国。 19世紀 （じゅうきゅうせいき）は、西暦1801年から西暦1900年までの100年間を指す世紀。 目次 1 19世紀の歴史 1.1 国民国家の成立 1.2 帝国主義の興隆 1.3 列強の植民地争奪戦 2 できごと 2.1 1800年代 2.2 1810年代 2.3 1820年代 2.4 1830年代 2.5 1840年代 2.6 1850年代 2.7 1860年代 2.8 1870年代 2.9 1880年代 2.10 1890年代 2.11 1900年代 3 文化 3.1 文学 3.2 音楽 3.3 思想 3.4 科学 3.5 技術 4 人物 4.1 ヨーロッパ 4.1.1 政治家・王族 4.1.1.1 フランス 4.1.1.2 オーストリア＝ハンガリー 4.1.1.3 ロシア 4.1.1.4 イギリス 4.1.1.5 ドイツ（プロイセンほかドイツ領邦を含む） 4.1.1.6 北欧 4.1.1.7 イタリア 4.1.1.8 スペイン 4.1.1.9 ベルギー 4.1.1.10 ギリシア 4.1.2 軍人 4.1.3 実業家 4.1.4 科学と技術 4.1.5 思想と哲学・人文諸学 4.1.6 宗教 4.1.7 文学 4.1.8 美術 4.1.9 音楽 4.1.10 社会事業家 4.1.11 探検家・旅行家 4.1.12 料
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