Convergence to a fixed point












3












$begingroup$


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



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



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



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






analysis convergence numerical-methods fixed-point-theorems fixedpoints







share|cite|improve this question









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


















3












$begingroup$


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



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



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



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






analysis convergence numerical-methods fixed-point-theorems fixedpoints







share|cite|improve this question









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
















3












3








3





$begingroup$


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



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



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



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






analysis convergence numerical-methods fixed-point-theorems fixedpoints







share|cite|improve this question









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.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







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 4 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




















  • $begingroup$
    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












3 Answers
3






active

oldest

votes


















2












$begingroup$

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






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    This can be proved using the Banach Fixed Point Theorem.



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



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



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



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



    This implies that



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



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



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






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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



















    1












    $begingroup$

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






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });






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










      draft saved

      draft discarded


















      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









      2












      $begingroup$

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






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

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






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

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






          share|cite|improve this answer











          $endgroup$



          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 4 hours ago

























          answered 4 hours ago









          Robert ShoreRobert Shore

          2,044116




          2,044116























              1












              $begingroup$

              This can be proved using the Banach Fixed Point Theorem.



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



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



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



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



              This implies that



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



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



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






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                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
















              1












              $begingroup$

              This can be proved using the Banach Fixed Point Theorem.



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



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



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



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



              This implies that



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



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



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






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                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














              1












              1








              1





              $begingroup$

              This can be proved using the Banach Fixed Point Theorem.



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



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



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



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



              This implies that



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



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



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






              share|cite|improve this answer









              $endgroup$



              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 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


















              • $begingroup$
                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











              1












              $begingroup$

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






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

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






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

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






                  share|cite|improve this answer









                  $endgroup$



                  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 3 hours ago









                  Kavi Rama MurthyKavi Rama Murthy

                  65k42766




                  65k42766






















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










                      draft saved

                      draft discarded


















                      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.




                      draft saved


                      draft discarded














                      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





















































                      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







                      -

                      Popular posts from this blog

                      What other Star Trek series did the main TNG cast show up in?

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 4 I'm referring to Picard, Riker, Data, Worf, Geordi, Dr. Crusher, Wesley Crusher, Chief O'Brien and family, Deanna Troi, Guinan. I know Picard and Worf show up on DS9, though I haven't seen it myself. I'm curious what happened to the characters after that last episode of TNG? star-trek star-trek-tng share | improve this question edited 10 hours ago Ham Sandwich 6,051 2 25 75 ...
                      Read more

                      Berlina muro

                      Image
                      La berlina muro ĉe Bethaniendamm, 1986 La berlina muro aŭ simple la muro [1] , nomita ankaŭ "kontraŭfaŝisma remparo" en propaganda lingvo de la iama Germana Demokratia Respubliko, estis parto de la interngermana limo, kiu dividis de la 13-a de aŭgusto 1961 ĝis la 9-a de novembro 1989 okcidentan Berlinon de la orienta parto de la urbo kaj de la ĉirkaŭa tereno de la GDR. Ĝi estis unu el la plej famaj simboloj de la malvarma milito kaj de la disdivido de Germanio. Pli ol simpla muro, la berlina muro estis fakte tuta defendokonstruaĵo kun du muroj, ena malplena strio, rondirvojo, gardturoj kaj alarmsistemoj. Dum la provo transiri la severe garditan limon al okcidenta Berlino multaj homoj estis mortigitaj. La preciza nombro de viktimoj estas pridisputita kaj ne certa; la nombroj anoncataj varias inter 86 kaj 238 mortintoj. La malplifortiĝo de Sovetunio kaj la politiko de liberigo gvidita de Miĥail Gorbaĉov ebligis al la orientgermanoj faligi la 9-an de novembro 1989 l...
                      Read more

                      Berlina aerponto

                      Image
                      Flugkoridoroj dum la blokado de Berlino La Berlina aerponto estas la provizigo de okcidenta Berlino per aviadiloj fare de la aliancanoj inter la 23-a de junio 1948 kaj la 12-a de majo 1949 dum la blokado de Berlino fare de la sovetia armeo. Enhavo 1 Historio 1.1 La aerkoridoroj 1.2 Unuaj flugoj 1.3 Plibonigo de la aerponto 1.4 Fino de la aerponto 1.5 Rozinbombiloj 1.6 Bilanco 2 Flughavenoj 2.1 Berlino 2.2 Okcidenta Germanio 3 Aviadiloj de la berlina aerponto 4 15-a datreveno de la aerponto 5 Bibliografio 6 Referencoj 7 Eksteraj ligiloj Historio | General Lucius D. Clay, militguberniestro de la usona okupadzono Jam antaŭ la berlina aerponto estiĝis la malgranda aerponto . La estro de la sovetia militadministracio en Germanio, Vasili Daniloviĉ Sokolovski, ordonis la 1-an de aprilo 1948, kiel reago al la konferenco de Londono inter britoj, francoj kaj reprezentantoj de Benelukso pri aligo de Germa...
                      Read more