Rate of convergence $x_{n+1}=2sin(x_n)$












0












$begingroup$


$f(x)=sin(x)-frac{1}{2}x$ , for $x>0$



We are trying to evaluate the root of function using the following fixed-point iterative method:



$x_{n+1}=2sin(x_n)$



What is the rate of convergence of the method?



Any direction would be appreciated






numerical-methods fixed-point-theorems







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    $f(x)=sin(x)-frac{1}{2}x$ , for $x>0$



    We are trying to evaluate the root of function using the following fixed-point iterative method:



    $x_{n+1}=2sin(x_n)$



    What is the rate of convergence of the method?



    Any direction would be appreciated






    numerical-methods fixed-point-theorems







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      $f(x)=sin(x)-frac{1}{2}x$ , for $x>0$



      We are trying to evaluate the root of function using the following fixed-point iterative method:



      $x_{n+1}=2sin(x_n)$



      What is the rate of convergence of the method?



      Any direction would be appreciated






      numerical-methods fixed-point-theorems







      share|cite|improve this question











      $endgroup$




      $f(x)=sin(x)-frac{1}{2}x$ , for $x>0$



      We are trying to evaluate the root of function using the following fixed-point iterative method:



      $x_{n+1}=2sin(x_n)$



      What is the rate of convergence of the method?



      Any direction would be appreciated






      numerical-methods fixed-point-theorems




      numerical-methods fixed-point-theorems






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 17:37







      Um Shmum

















      asked Jan 8 at 17:04









      Um ShmumUm Shmum

      1298




      1298






















          1 Answer
          1






          active

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          2












          $begingroup$

          You should easily see that $x=0$ is an instable fixed point. Then also that $g([0,pi])subset[0,2]subset[0,pi]$ so that there is an attracting fixed point inside that interval.



          Now you need to find better bounds for that point so that you can then compute a contraction factor as the maximum of the absolute value of the derivative. For instance, it is easy to see that $g([fracpi3,frac{2pi}3])=[sqrt3,2]subset [fracpi3,frac{2pi}3]$ and $2cos([fracpi3,frac{2pi}3])subset[-1,1]$, which is not sufficient but still closer.




          Numerically you get
          $$
          g([1.85, 1.95])subset[1.85, 1.95]
          ~~text{ with }~~
          g'([1.85, 1.95])subset[-0.75, -0.55]
          $$
          so that $frac34$ is an upper bound for the convergence rate, $frac12$ a lower bound.




          One might get analytical results using
          $$
          g(tfracpi2+u)=2cos(u)in[2-u^2, 2-u^2+tfrac1{12}u^4]subset [2-u^2, 2-tfrac{47}{48}u^2] ~~text{ for }~~ uin[0,tfrac12].
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do you prove there is an attracting fixed point?
            $endgroup$
            – lhf
            Jan 8 at 17:29












          • $begingroup$
            How does it help me compute the rate of convergence?
            $endgroup$
            – Um Shmum
            Jan 8 at 17:38










          • $begingroup$
            @lhf : Because the orbits starting anywhere in $(0,pi]$ have to end somewhere, and that "somewhere" can not be the point $0$. But to make that precise one needs the mentioned better bounds or smaller interval with a contraction constant to exclude the more exotic scenarios of limit cycles etc.
            $endgroup$
            – LutzL
            Jan 8 at 17:39












          • $begingroup$
            @UmShmum : You can only expect linear order of convergence. The contraction constant is an upper bound for the convergence rate. It is also a good approximation, if the bounding interval is sufficiently small.
            $endgroup$
            – LutzL
            Jan 8 at 17:42











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          1 Answer
          1






          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          You should easily see that $x=0$ is an instable fixed point. Then also that $g([0,pi])subset[0,2]subset[0,pi]$ so that there is an attracting fixed point inside that interval.



          Now you need to find better bounds for that point so that you can then compute a contraction factor as the maximum of the absolute value of the derivative. For instance, it is easy to see that $g([fracpi3,frac{2pi}3])=[sqrt3,2]subset [fracpi3,frac{2pi}3]$ and $2cos([fracpi3,frac{2pi}3])subset[-1,1]$, which is not sufficient but still closer.




          Numerically you get
          $$
          g([1.85, 1.95])subset[1.85, 1.95]
          ~~text{ with }~~
          g'([1.85, 1.95])subset[-0.75, -0.55]
          $$
          so that $frac34$ is an upper bound for the convergence rate, $frac12$ a lower bound.




          One might get analytical results using
          $$
          g(tfracpi2+u)=2cos(u)in[2-u^2, 2-u^2+tfrac1{12}u^4]subset [2-u^2, 2-tfrac{47}{48}u^2] ~~text{ for }~~ uin[0,tfrac12].
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do you prove there is an attracting fixed point?
            $endgroup$
            – lhf
            Jan 8 at 17:29












          • $begingroup$
            How does it help me compute the rate of convergence?
            $endgroup$
            – Um Shmum
            Jan 8 at 17:38










          • $begingroup$
            @lhf : Because the orbits starting anywhere in $(0,pi]$ have to end somewhere, and that "somewhere" can not be the point $0$. But to make that precise one needs the mentioned better bounds or smaller interval with a contraction constant to exclude the more exotic scenarios of limit cycles etc.
            $endgroup$
            – LutzL
            Jan 8 at 17:39












          • $begingroup$
            @UmShmum : You can only expect linear order of convergence. The contraction constant is an upper bound for the convergence rate. It is also a good approximation, if the bounding interval is sufficiently small.
            $endgroup$
            – LutzL
            Jan 8 at 17:42
















          2












          $begingroup$

          You should easily see that $x=0$ is an instable fixed point. Then also that $g([0,pi])subset[0,2]subset[0,pi]$ so that there is an attracting fixed point inside that interval.



          Now you need to find better bounds for that point so that you can then compute a contraction factor as the maximum of the absolute value of the derivative. For instance, it is easy to see that $g([fracpi3,frac{2pi}3])=[sqrt3,2]subset [fracpi3,frac{2pi}3]$ and $2cos([fracpi3,frac{2pi}3])subset[-1,1]$, which is not sufficient but still closer.




          Numerically you get
          $$
          g([1.85, 1.95])subset[1.85, 1.95]
          ~~text{ with }~~
          g'([1.85, 1.95])subset[-0.75, -0.55]
          $$
          so that $frac34$ is an upper bound for the convergence rate, $frac12$ a lower bound.




          One might get analytical results using
          $$
          g(tfracpi2+u)=2cos(u)in[2-u^2, 2-u^2+tfrac1{12}u^4]subset [2-u^2, 2-tfrac{47}{48}u^2] ~~text{ for }~~ uin[0,tfrac12].
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How do you prove there is an attracting fixed point?
            $endgroup$
            – lhf
            Jan 8 at 17:29












          • $begingroup$
            How does it help me compute the rate of convergence?
            $endgroup$
            – Um Shmum
            Jan 8 at 17:38










          • $begingroup$
            @lhf : Because the orbits starting anywhere in $(0,pi]$ have to end somewhere, and that "somewhere" can not be the point $0$. But to make that precise one needs the mentioned better bounds or smaller interval with a contraction constant to exclude the more exotic scenarios of limit cycles etc.
            $endgroup$
            – LutzL
            Jan 8 at 17:39












          • $begingroup$
            @UmShmum : You can only expect linear order of convergence. The contraction constant is an upper bound for the convergence rate. It is also a good approximation, if the bounding interval is sufficiently small.
            $endgroup$
            – LutzL
            Jan 8 at 17:42














          2












          2








          2





          $begingroup$

          You should easily see that $x=0$ is an instable fixed point. Then also that $g([0,pi])subset[0,2]subset[0,pi]$ so that there is an attracting fixed point inside that interval.



          Now you need to find better bounds for that point so that you can then compute a contraction factor as the maximum of the absolute value of the derivative. For instance, it is easy to see that $g([fracpi3,frac{2pi}3])=[sqrt3,2]subset [fracpi3,frac{2pi}3]$ and $2cos([fracpi3,frac{2pi}3])subset[-1,1]$, which is not sufficient but still closer.




          Numerically you get
          $$
          g([1.85, 1.95])subset[1.85, 1.95]
          ~~text{ with }~~
          g'([1.85, 1.95])subset[-0.75, -0.55]
          $$
          so that $frac34$ is an upper bound for the convergence rate, $frac12$ a lower bound.




          One might get analytical results using
          $$
          g(tfracpi2+u)=2cos(u)in[2-u^2, 2-u^2+tfrac1{12}u^4]subset [2-u^2, 2-tfrac{47}{48}u^2] ~~text{ for }~~ uin[0,tfrac12].
          $$






          share|cite|improve this answer











          $endgroup$



          You should easily see that $x=0$ is an instable fixed point. Then also that $g([0,pi])subset[0,2]subset[0,pi]$ so that there is an attracting fixed point inside that interval.



          Now you need to find better bounds for that point so that you can then compute a contraction factor as the maximum of the absolute value of the derivative. For instance, it is easy to see that $g([fracpi3,frac{2pi}3])=[sqrt3,2]subset [fracpi3,frac{2pi}3]$ and $2cos([fracpi3,frac{2pi}3])subset[-1,1]$, which is not sufficient but still closer.




          Numerically you get
          $$
          g([1.85, 1.95])subset[1.85, 1.95]
          ~~text{ with }~~
          g'([1.85, 1.95])subset[-0.75, -0.55]
          $$
          so that $frac34$ is an upper bound for the convergence rate, $frac12$ a lower bound.




          One might get analytical results using
          $$
          g(tfracpi2+u)=2cos(u)in[2-u^2, 2-u^2+tfrac1{12}u^4]subset [2-u^2, 2-tfrac{47}{48}u^2] ~~text{ for }~~ uin[0,tfrac12].
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 23:03

























          answered Jan 8 at 17:24









          LutzLLutzL

          57.4k42054




          57.4k42054












          • $begingroup$
            How do you prove there is an attracting fixed point?
            $endgroup$
            – lhf
            Jan 8 at 17:29












          • $begingroup$
            How does it help me compute the rate of convergence?
            $endgroup$
            – Um Shmum
            Jan 8 at 17:38










          • $begingroup$
            @lhf : Because the orbits starting anywhere in $(0,pi]$ have to end somewhere, and that "somewhere" can not be the point $0$. But to make that precise one needs the mentioned better bounds or smaller interval with a contraction constant to exclude the more exotic scenarios of limit cycles etc.
            $endgroup$
            – LutzL
            Jan 8 at 17:39












          • $begingroup$
            @UmShmum : You can only expect linear order of convergence. The contraction constant is an upper bound for the convergence rate. It is also a good approximation, if the bounding interval is sufficiently small.
            $endgroup$
            – LutzL
            Jan 8 at 17:42


















          • $begingroup$
            How do you prove there is an attracting fixed point?
            $endgroup$
            – lhf
            Jan 8 at 17:29












          • $begingroup$
            How does it help me compute the rate of convergence?
            $endgroup$
            – Um Shmum
            Jan 8 at 17:38










          • $begingroup$
            @lhf : Because the orbits starting anywhere in $(0,pi]$ have to end somewhere, and that "somewhere" can not be the point $0$. But to make that precise one needs the mentioned better bounds or smaller interval with a contraction constant to exclude the more exotic scenarios of limit cycles etc.
            $endgroup$
            – LutzL
            Jan 8 at 17:39












          • $begingroup$
            @UmShmum : You can only expect linear order of convergence. The contraction constant is an upper bound for the convergence rate. It is also a good approximation, if the bounding interval is sufficiently small.
            $endgroup$
            – LutzL
            Jan 8 at 17:42
















          $begingroup$
          How do you prove there is an attracting fixed point?
          $endgroup$
          – lhf
          Jan 8 at 17:29






          $begingroup$
          How do you prove there is an attracting fixed point?
          $endgroup$
          – lhf
          Jan 8 at 17:29














          $begingroup$
          How does it help me compute the rate of convergence?
          $endgroup$
          – Um Shmum
          Jan 8 at 17:38




          $begingroup$
          How does it help me compute the rate of convergence?
          $endgroup$
          – Um Shmum
          Jan 8 at 17:38












          $begingroup$
          @lhf : Because the orbits starting anywhere in $(0,pi]$ have to end somewhere, and that "somewhere" can not be the point $0$. But to make that precise one needs the mentioned better bounds or smaller interval with a contraction constant to exclude the more exotic scenarios of limit cycles etc.
          $endgroup$
          – LutzL
          Jan 8 at 17:39






          $begingroup$
          @lhf : Because the orbits starting anywhere in $(0,pi]$ have to end somewhere, and that "somewhere" can not be the point $0$. But to make that precise one needs the mentioned better bounds or smaller interval with a contraction constant to exclude the more exotic scenarios of limit cycles etc.
          $endgroup$
          – LutzL
          Jan 8 at 17:39














          $begingroup$
          @UmShmum : You can only expect linear order of convergence. The contraction constant is an upper bound for the convergence rate. It is also a good approximation, if the bounding interval is sufficiently small.
          $endgroup$
          – LutzL
          Jan 8 at 17:42




          $begingroup$
          @UmShmum : You can only expect linear order of convergence. The contraction constant is an upper bound for the convergence rate. It is also a good approximation, if the bounding interval is sufficiently small.
          $endgroup$
          – LutzL
          Jan 8 at 17:42


















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