Iteration for fixed point












3












$begingroup$


Suppose $x_{k+1}= g(x_k)$ is fixed point iteration for some continuously diffrentiable $g(x)$. The theorem im learning says that if $g(r) = r$ and $|g'(r)| < 1$ then the fixed point iteration converges to $r$ for initial guess $x_0$ sufficiently close to $r$.



MY question is: Is the converse is also true? That is, if the fixed point iteration converges to $r$, then we must have $|g'(r)|<1$?



OR is it possible to have situations where $g'(r) geq 1$ with $(x_k)$ convergent to $r$.






numerical-methods







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$endgroup$

















    3












    $begingroup$


    Suppose $x_{k+1}= g(x_k)$ is fixed point iteration for some continuously diffrentiable $g(x)$. The theorem im learning says that if $g(r) = r$ and $|g'(r)| < 1$ then the fixed point iteration converges to $r$ for initial guess $x_0$ sufficiently close to $r$.



    MY question is: Is the converse is also true? That is, if the fixed point iteration converges to $r$, then we must have $|g'(r)|<1$?



    OR is it possible to have situations where $g'(r) geq 1$ with $(x_k)$ convergent to $r$.






    numerical-methods







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Suppose $x_{k+1}= g(x_k)$ is fixed point iteration for some continuously diffrentiable $g(x)$. The theorem im learning says that if $g(r) = r$ and $|g'(r)| < 1$ then the fixed point iteration converges to $r$ for initial guess $x_0$ sufficiently close to $r$.



      MY question is: Is the converse is also true? That is, if the fixed point iteration converges to $r$, then we must have $|g'(r)|<1$?



      OR is it possible to have situations where $g'(r) geq 1$ with $(x_k)$ convergent to $r$.






      numerical-methods







      share|cite|improve this question









      $endgroup$




      Suppose $x_{k+1}= g(x_k)$ is fixed point iteration for some continuously diffrentiable $g(x)$. The theorem im learning says that if $g(r) = r$ and $|g'(r)| < 1$ then the fixed point iteration converges to $r$ for initial guess $x_0$ sufficiently close to $r$.



      MY question is: Is the converse is also true? That is, if the fixed point iteration converges to $r$, then we must have $|g'(r)|<1$?



      OR is it possible to have situations where $g'(r) geq 1$ with $(x_k)$ convergent to $r$.






      numerical-methods




      numerical-methods






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 8 hours ago









      Jimmy SabaterJimmy Sabater

      2,240319




      2,240319






















          3 Answers
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          active

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          2












          $begingroup$

          Let $g(x)=x$. Then the sequence $(x_k)$ is constant hence convergent. We have $g'(x)=1$ for all $x$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            how about if $|g'(r)| > 1$ can we find counterexample ?
            $endgroup$
            – Jimmy Sabater
            8 hours ago



















          2












          $begingroup$

          Suppose you have the function $f(x)=kx(x-a)+a$ so that $f(a)=a$



          Then $f'(x)=2kx-k = k(2x-1)$ and if $aneq frac 12$ it is possible to choose a value of $k$ to make $f'(a)$ any value you choose.



          The difference is that if $|f'(a)| gt 1$ there is no open interval containing $a$ in which the iteration converges - this only happens at the point.



          The behaviour in general where $|f'(a)| = 1$ depends on the function.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            The iteration $x_{n+1}=sin(x_n)$ converges towards $r=0$ despite the derivative there being $cos(0)=1$.




            For $y_k=x_k^2$ one has the estimate by the Leibniz rule on alternating series
            $$small
            y_{k+1}=frac12(1-cos(2x_k))
            le y_k-frac13y_k^2+frac2{45}y_k^3
            %=y_kfrac{1-frac{1}{15}y_k^2-frac2{135}y_k^3}{1+frac13y_k}
            lefrac{y_k}{1+frac13y_k}\~\small
            implies y_{k+1}^{-1}gefrac13+y_k^{-1}implies y_klefrac{y_0}{1+frac{k}3y_0}
            $$
            so that one finds the convergence by the non-geometric majorant
            $$small
            |x_k|lefrac{|x_0|}{sqrt{1+frac{k}3x_0^2}}.
            $$






            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

              votes









              2












              $begingroup$

              Let $g(x)=x$. Then the sequence $(x_k)$ is constant hence convergent. We have $g'(x)=1$ for all $x$.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                how about if $|g'(r)| > 1$ can we find counterexample ?
                $endgroup$
                – Jimmy Sabater
                8 hours ago
















              2












              $begingroup$

              Let $g(x)=x$. Then the sequence $(x_k)$ is constant hence convergent. We have $g'(x)=1$ for all $x$.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                how about if $|g'(r)| > 1$ can we find counterexample ?
                $endgroup$
                – Jimmy Sabater
                8 hours ago














              2












              2








              2





              $begingroup$

              Let $g(x)=x$. Then the sequence $(x_k)$ is constant hence convergent. We have $g'(x)=1$ for all $x$.






              share|cite|improve this answer









              $endgroup$



              Let $g(x)=x$. Then the sequence $(x_k)$ is constant hence convergent. We have $g'(x)=1$ for all $x$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 8 hours ago









              FredFred

              44.7k1846




              44.7k1846








              • 1




                $begingroup$
                how about if $|g'(r)| > 1$ can we find counterexample ?
                $endgroup$
                – Jimmy Sabater
                8 hours ago














              • 1




                $begingroup$
                how about if $|g'(r)| > 1$ can we find counterexample ?
                $endgroup$
                – Jimmy Sabater
                8 hours ago








              1




              1




              $begingroup$
              how about if $|g'(r)| > 1$ can we find counterexample ?
              $endgroup$
              – Jimmy Sabater
              8 hours ago




              $begingroup$
              how about if $|g'(r)| > 1$ can we find counterexample ?
              $endgroup$
              – Jimmy Sabater
              8 hours ago











              2












              $begingroup$

              Suppose you have the function $f(x)=kx(x-a)+a$ so that $f(a)=a$



              Then $f'(x)=2kx-k = k(2x-1)$ and if $aneq frac 12$ it is possible to choose a value of $k$ to make $f'(a)$ any value you choose.



              The difference is that if $|f'(a)| gt 1$ there is no open interval containing $a$ in which the iteration converges - this only happens at the point.



              The behaviour in general where $|f'(a)| = 1$ depends on the function.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Suppose you have the function $f(x)=kx(x-a)+a$ so that $f(a)=a$



                Then $f'(x)=2kx-k = k(2x-1)$ and if $aneq frac 12$ it is possible to choose a value of $k$ to make $f'(a)$ any value you choose.



                The difference is that if $|f'(a)| gt 1$ there is no open interval containing $a$ in which the iteration converges - this only happens at the point.



                The behaviour in general where $|f'(a)| = 1$ depends on the function.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Suppose you have the function $f(x)=kx(x-a)+a$ so that $f(a)=a$



                  Then $f'(x)=2kx-k = k(2x-1)$ and if $aneq frac 12$ it is possible to choose a value of $k$ to make $f'(a)$ any value you choose.



                  The difference is that if $|f'(a)| gt 1$ there is no open interval containing $a$ in which the iteration converges - this only happens at the point.



                  The behaviour in general where $|f'(a)| = 1$ depends on the function.






                  share|cite|improve this answer









                  $endgroup$



                  Suppose you have the function $f(x)=kx(x-a)+a$ so that $f(a)=a$



                  Then $f'(x)=2kx-k = k(2x-1)$ and if $aneq frac 12$ it is possible to choose a value of $k$ to make $f'(a)$ any value you choose.



                  The difference is that if $|f'(a)| gt 1$ there is no open interval containing $a$ in which the iteration converges - this only happens at the point.



                  The behaviour in general where $|f'(a)| = 1$ depends on the function.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 8 hours ago









                  Mark BennetMark Bennet

                  80.9k981179




                  80.9k981179























                      2












                      $begingroup$

                      The iteration $x_{n+1}=sin(x_n)$ converges towards $r=0$ despite the derivative there being $cos(0)=1$.




                      For $y_k=x_k^2$ one has the estimate by the Leibniz rule on alternating series
                      $$small
                      y_{k+1}=frac12(1-cos(2x_k))
                      le y_k-frac13y_k^2+frac2{45}y_k^3
                      %=y_kfrac{1-frac{1}{15}y_k^2-frac2{135}y_k^3}{1+frac13y_k}
                      lefrac{y_k}{1+frac13y_k}\~\small
                      implies y_{k+1}^{-1}gefrac13+y_k^{-1}implies y_klefrac{y_0}{1+frac{k}3y_0}
                      $$
                      so that one finds the convergence by the non-geometric majorant
                      $$small
                      |x_k|lefrac{|x_0|}{sqrt{1+frac{k}3x_0^2}}.
                      $$






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        The iteration $x_{n+1}=sin(x_n)$ converges towards $r=0$ despite the derivative there being $cos(0)=1$.




                        For $y_k=x_k^2$ one has the estimate by the Leibniz rule on alternating series
                        $$small
                        y_{k+1}=frac12(1-cos(2x_k))
                        le y_k-frac13y_k^2+frac2{45}y_k^3
                        %=y_kfrac{1-frac{1}{15}y_k^2-frac2{135}y_k^3}{1+frac13y_k}
                        lefrac{y_k}{1+frac13y_k}\~\small
                        implies y_{k+1}^{-1}gefrac13+y_k^{-1}implies y_klefrac{y_0}{1+frac{k}3y_0}
                        $$
                        so that one finds the convergence by the non-geometric majorant
                        $$small
                        |x_k|lefrac{|x_0|}{sqrt{1+frac{k}3x_0^2}}.
                        $$






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          The iteration $x_{n+1}=sin(x_n)$ converges towards $r=0$ despite the derivative there being $cos(0)=1$.




                          For $y_k=x_k^2$ one has the estimate by the Leibniz rule on alternating series
                          $$small
                          y_{k+1}=frac12(1-cos(2x_k))
                          le y_k-frac13y_k^2+frac2{45}y_k^3
                          %=y_kfrac{1-frac{1}{15}y_k^2-frac2{135}y_k^3}{1+frac13y_k}
                          lefrac{y_k}{1+frac13y_k}\~\small
                          implies y_{k+1}^{-1}gefrac13+y_k^{-1}implies y_klefrac{y_0}{1+frac{k}3y_0}
                          $$
                          so that one finds the convergence by the non-geometric majorant
                          $$small
                          |x_k|lefrac{|x_0|}{sqrt{1+frac{k}3x_0^2}}.
                          $$






                          share|cite|improve this answer









                          $endgroup$



                          The iteration $x_{n+1}=sin(x_n)$ converges towards $r=0$ despite the derivative there being $cos(0)=1$.




                          For $y_k=x_k^2$ one has the estimate by the Leibniz rule on alternating series
                          $$small
                          y_{k+1}=frac12(1-cos(2x_k))
                          le y_k-frac13y_k^2+frac2{45}y_k^3
                          %=y_kfrac{1-frac{1}{15}y_k^2-frac2{135}y_k^3}{1+frac13y_k}
                          lefrac{y_k}{1+frac13y_k}\~\small
                          implies y_{k+1}^{-1}gefrac13+y_k^{-1}implies y_klefrac{y_0}{1+frac{k}3y_0}
                          $$
                          so that one finds the convergence by the non-geometric majorant
                          $$small
                          |x_k|lefrac{|x_0|}{sqrt{1+frac{k}3x_0^2}}.
                          $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 5 hours ago









                          LutzLLutzL

                          57.3k42054




                          57.3k42054






























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