Order of convergence of a two-step iteration method












0












$begingroup$


Find the order of the iteration is given by



$$y_{n+1} = x_n - frac{f(x_n)}{f'(x_n)} quad x_{n+1} = y_{n+1} - frac{f(y_{n+1})}{f'(x_n)}$$



Assuming $f(r) = 0$, $f'(r) neq 0$ and the initial guess is close to r.



I know the iteration is a third order method because if I let $Y(x) = x - frac{f(x)}{f'(x)}$ and $I(x) = Y(x) - frac{f(Y(x))}{f'(x)}$. We end up with
$$frac{dI(x)}{dx}|_r = 0$$
$$frac{d^2I(x)}{dx^2}|_r = 0$$
$$frac{d^3I(x)}{dx^3}|_r neq 0$$



This way however is very messy when evaluating all of the derivatives. I was wondering if there is another way to prove that the above methods order of convergence?






numerical-methods







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Find the order of the iteration is given by



    $$y_{n+1} = x_n - frac{f(x_n)}{f'(x_n)} quad x_{n+1} = y_{n+1} - frac{f(y_{n+1})}{f'(x_n)}$$



    Assuming $f(r) = 0$, $f'(r) neq 0$ and the initial guess is close to r.



    I know the iteration is a third order method because if I let $Y(x) = x - frac{f(x)}{f'(x)}$ and $I(x) = Y(x) - frac{f(Y(x))}{f'(x)}$. We end up with
    $$frac{dI(x)}{dx}|_r = 0$$
    $$frac{d^2I(x)}{dx^2}|_r = 0$$
    $$frac{d^3I(x)}{dx^3}|_r neq 0$$



    This way however is very messy when evaluating all of the derivatives. I was wondering if there is another way to prove that the above methods order of convergence?






    numerical-methods







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Find the order of the iteration is given by



      $$y_{n+1} = x_n - frac{f(x_n)}{f'(x_n)} quad x_{n+1} = y_{n+1} - frac{f(y_{n+1})}{f'(x_n)}$$



      Assuming $f(r) = 0$, $f'(r) neq 0$ and the initial guess is close to r.



      I know the iteration is a third order method because if I let $Y(x) = x - frac{f(x)}{f'(x)}$ and $I(x) = Y(x) - frac{f(Y(x))}{f'(x)}$. We end up with
      $$frac{dI(x)}{dx}|_r = 0$$
      $$frac{d^2I(x)}{dx^2}|_r = 0$$
      $$frac{d^3I(x)}{dx^3}|_r neq 0$$



      This way however is very messy when evaluating all of the derivatives. I was wondering if there is another way to prove that the above methods order of convergence?






      numerical-methods







      share|cite|improve this question











      $endgroup$




      Find the order of the iteration is given by



      $$y_{n+1} = x_n - frac{f(x_n)}{f'(x_n)} quad x_{n+1} = y_{n+1} - frac{f(y_{n+1})}{f'(x_n)}$$



      Assuming $f(r) = 0$, $f'(r) neq 0$ and the initial guess is close to r.



      I know the iteration is a third order method because if I let $Y(x) = x - frac{f(x)}{f'(x)}$ and $I(x) = Y(x) - frac{f(Y(x))}{f'(x)}$. We end up with
      $$frac{dI(x)}{dx}|_r = 0$$
      $$frac{d^2I(x)}{dx^2}|_r = 0$$
      $$frac{d^3I(x)}{dx^3}|_r neq 0$$



      This way however is very messy when evaluating all of the derivatives. I was wondering if there is another way to prove that the above methods order of convergence?






      numerical-methods




      numerical-methods






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 19:44









      Rebellos

      15.3k31250




      15.3k31250










      asked Dec 16 '18 at 19:40







      user627004





























          1 Answer
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          $begingroup$

          Let $L$ be a Lipschitz constant of $f'$ on the domain under consideration. Then we know that
          $$
          |f(x+v)-f(x)-f'(x)v|leint_0^1|f'(x+sv)-f'(x)|,|v|,dsle L,|v|,int_0^1|sv|,ds=frac{L}2|v|^2,
          $$
          so that especially
          $$
          |f(y_{n+1})|lefrac{L}2frac{|f(x_n)|^2}{|f'(x_n)|^2}
          $$
          Further,
          $$
          f(x_{n+1})=f(y_{n+1})-f'(y_{n+1})frac{f(y_{n+1})}{f'(x_n)}+R~~text{with}~~ |R|lefrac{L}2frac{|f(y_{n+1}|)^2}{|f'(x_n)|^2}
          $$
          so that
          $$
          |f(x_{n+1})|
          le L,|x_{n+1}-y_{n+1}|,frac{|f(y_{n+1})|}{|f'(x_n)|}+frac{L^3}{8}frac{|f(x_{n+1})|^4}{|f'(x_n)|^6}
          le frac{L^2}{2}frac{|f(x_n)|^3}{|f'(x_n)|^4}+frac{L^3}{8}frac{|f(x_{n+1})|^4}{|f'(x_n)|^6}
          $$
          As the function value stands as proxy for the distance to the root, this shows the third order convergence.




          Note however that in terms of effort, or the Ostrowski index, you only get for polynomial $f$ an order of $sqrt[3]{3}=1.442$ per function and derivative evaluation, in the non-polynomial case even only $sqrt[4]3=1.316$ as one derivative costs as much as 2 function evaluations. To compare, the Newton method has $sqrt2=1.414$ resp. $sqrt[3]2=1.260$ and the secant method $frac{1+sqrt5}2=1.618$ (if it converges).



          So this method is slightly better than the Newton method. To put it in integer terms, in the polynomial case within the effort of $6$ function evaluations one can perform




          • 3 Newton steps with error reduction from $epsilon$ to $ϵ^8$ or

          • 2 cycles of this two-step method with an error reduction to $ϵ^9$.


          In the non-polynomial case, in the time of $12$ evaluations of $f$ one can perform





          • $4$ Newton steps to $ϵ^{16}$ or


          • $3$ two-step cycles to $ϵ^{27}$.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            Let $L$ be a Lipschitz constant of $f'$ on the domain under consideration. Then we know that
            $$
            |f(x+v)-f(x)-f'(x)v|leint_0^1|f'(x+sv)-f'(x)|,|v|,dsle L,|v|,int_0^1|sv|,ds=frac{L}2|v|^2,
            $$
            so that especially
            $$
            |f(y_{n+1})|lefrac{L}2frac{|f(x_n)|^2}{|f'(x_n)|^2}
            $$
            Further,
            $$
            f(x_{n+1})=f(y_{n+1})-f'(y_{n+1})frac{f(y_{n+1})}{f'(x_n)}+R~~text{with}~~ |R|lefrac{L}2frac{|f(y_{n+1}|)^2}{|f'(x_n)|^2}
            $$
            so that
            $$
            |f(x_{n+1})|
            le L,|x_{n+1}-y_{n+1}|,frac{|f(y_{n+1})|}{|f'(x_n)|}+frac{L^3}{8}frac{|f(x_{n+1})|^4}{|f'(x_n)|^6}
            le frac{L^2}{2}frac{|f(x_n)|^3}{|f'(x_n)|^4}+frac{L^3}{8}frac{|f(x_{n+1})|^4}{|f'(x_n)|^6}
            $$
            As the function value stands as proxy for the distance to the root, this shows the third order convergence.




            Note however that in terms of effort, or the Ostrowski index, you only get for polynomial $f$ an order of $sqrt[3]{3}=1.442$ per function and derivative evaluation, in the non-polynomial case even only $sqrt[4]3=1.316$ as one derivative costs as much as 2 function evaluations. To compare, the Newton method has $sqrt2=1.414$ resp. $sqrt[3]2=1.260$ and the secant method $frac{1+sqrt5}2=1.618$ (if it converges).



            So this method is slightly better than the Newton method. To put it in integer terms, in the polynomial case within the effort of $6$ function evaluations one can perform




            • 3 Newton steps with error reduction from $epsilon$ to $ϵ^8$ or

            • 2 cycles of this two-step method with an error reduction to $ϵ^9$.


            In the non-polynomial case, in the time of $12$ evaluations of $f$ one can perform





            • $4$ Newton steps to $ϵ^{16}$ or


            • $3$ two-step cycles to $ϵ^{27}$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Let $L$ be a Lipschitz constant of $f'$ on the domain under consideration. Then we know that
              $$
              |f(x+v)-f(x)-f'(x)v|leint_0^1|f'(x+sv)-f'(x)|,|v|,dsle L,|v|,int_0^1|sv|,ds=frac{L}2|v|^2,
              $$
              so that especially
              $$
              |f(y_{n+1})|lefrac{L}2frac{|f(x_n)|^2}{|f'(x_n)|^2}
              $$
              Further,
              $$
              f(x_{n+1})=f(y_{n+1})-f'(y_{n+1})frac{f(y_{n+1})}{f'(x_n)}+R~~text{with}~~ |R|lefrac{L}2frac{|f(y_{n+1}|)^2}{|f'(x_n)|^2}
              $$
              so that
              $$
              |f(x_{n+1})|
              le L,|x_{n+1}-y_{n+1}|,frac{|f(y_{n+1})|}{|f'(x_n)|}+frac{L^3}{8}frac{|f(x_{n+1})|^4}{|f'(x_n)|^6}
              le frac{L^2}{2}frac{|f(x_n)|^3}{|f'(x_n)|^4}+frac{L^3}{8}frac{|f(x_{n+1})|^4}{|f'(x_n)|^6}
              $$
              As the function value stands as proxy for the distance to the root, this shows the third order convergence.




              Note however that in terms of effort, or the Ostrowski index, you only get for polynomial $f$ an order of $sqrt[3]{3}=1.442$ per function and derivative evaluation, in the non-polynomial case even only $sqrt[4]3=1.316$ as one derivative costs as much as 2 function evaluations. To compare, the Newton method has $sqrt2=1.414$ resp. $sqrt[3]2=1.260$ and the secant method $frac{1+sqrt5}2=1.618$ (if it converges).



              So this method is slightly better than the Newton method. To put it in integer terms, in the polynomial case within the effort of $6$ function evaluations one can perform




              • 3 Newton steps with error reduction from $epsilon$ to $ϵ^8$ or

              • 2 cycles of this two-step method with an error reduction to $ϵ^9$.


              In the non-polynomial case, in the time of $12$ evaluations of $f$ one can perform





              • $4$ Newton steps to $ϵ^{16}$ or


              • $3$ two-step cycles to $ϵ^{27}$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Let $L$ be a Lipschitz constant of $f'$ on the domain under consideration. Then we know that
                $$
                |f(x+v)-f(x)-f'(x)v|leint_0^1|f'(x+sv)-f'(x)|,|v|,dsle L,|v|,int_0^1|sv|,ds=frac{L}2|v|^2,
                $$
                so that especially
                $$
                |f(y_{n+1})|lefrac{L}2frac{|f(x_n)|^2}{|f'(x_n)|^2}
                $$
                Further,
                $$
                f(x_{n+1})=f(y_{n+1})-f'(y_{n+1})frac{f(y_{n+1})}{f'(x_n)}+R~~text{with}~~ |R|lefrac{L}2frac{|f(y_{n+1}|)^2}{|f'(x_n)|^2}
                $$
                so that
                $$
                |f(x_{n+1})|
                le L,|x_{n+1}-y_{n+1}|,frac{|f(y_{n+1})|}{|f'(x_n)|}+frac{L^3}{8}frac{|f(x_{n+1})|^4}{|f'(x_n)|^6}
                le frac{L^2}{2}frac{|f(x_n)|^3}{|f'(x_n)|^4}+frac{L^3}{8}frac{|f(x_{n+1})|^4}{|f'(x_n)|^6}
                $$
                As the function value stands as proxy for the distance to the root, this shows the third order convergence.




                Note however that in terms of effort, or the Ostrowski index, you only get for polynomial $f$ an order of $sqrt[3]{3}=1.442$ per function and derivative evaluation, in the non-polynomial case even only $sqrt[4]3=1.316$ as one derivative costs as much as 2 function evaluations. To compare, the Newton method has $sqrt2=1.414$ resp. $sqrt[3]2=1.260$ and the secant method $frac{1+sqrt5}2=1.618$ (if it converges).



                So this method is slightly better than the Newton method. To put it in integer terms, in the polynomial case within the effort of $6$ function evaluations one can perform




                • 3 Newton steps with error reduction from $epsilon$ to $ϵ^8$ or

                • 2 cycles of this two-step method with an error reduction to $ϵ^9$.


                In the non-polynomial case, in the time of $12$ evaluations of $f$ one can perform





                • $4$ Newton steps to $ϵ^{16}$ or


                • $3$ two-step cycles to $ϵ^{27}$.






                share|cite|improve this answer











                $endgroup$



                Let $L$ be a Lipschitz constant of $f'$ on the domain under consideration. Then we know that
                $$
                |f(x+v)-f(x)-f'(x)v|leint_0^1|f'(x+sv)-f'(x)|,|v|,dsle L,|v|,int_0^1|sv|,ds=frac{L}2|v|^2,
                $$
                so that especially
                $$
                |f(y_{n+1})|lefrac{L}2frac{|f(x_n)|^2}{|f'(x_n)|^2}
                $$
                Further,
                $$
                f(x_{n+1})=f(y_{n+1})-f'(y_{n+1})frac{f(y_{n+1})}{f'(x_n)}+R~~text{with}~~ |R|lefrac{L}2frac{|f(y_{n+1}|)^2}{|f'(x_n)|^2}
                $$
                so that
                $$
                |f(x_{n+1})|
                le L,|x_{n+1}-y_{n+1}|,frac{|f(y_{n+1})|}{|f'(x_n)|}+frac{L^3}{8}frac{|f(x_{n+1})|^4}{|f'(x_n)|^6}
                le frac{L^2}{2}frac{|f(x_n)|^3}{|f'(x_n)|^4}+frac{L^3}{8}frac{|f(x_{n+1})|^4}{|f'(x_n)|^6}
                $$
                As the function value stands as proxy for the distance to the root, this shows the third order convergence.




                Note however that in terms of effort, or the Ostrowski index, you only get for polynomial $f$ an order of $sqrt[3]{3}=1.442$ per function and derivative evaluation, in the non-polynomial case even only $sqrt[4]3=1.316$ as one derivative costs as much as 2 function evaluations. To compare, the Newton method has $sqrt2=1.414$ resp. $sqrt[3]2=1.260$ and the secant method $frac{1+sqrt5}2=1.618$ (if it converges).



                So this method is slightly better than the Newton method. To put it in integer terms, in the polynomial case within the effort of $6$ function evaluations one can perform




                • 3 Newton steps with error reduction from $epsilon$ to $ϵ^8$ or

                • 2 cycles of this two-step method with an error reduction to $ϵ^9$.


                In the non-polynomial case, in the time of $12$ evaluations of $f$ one can perform





                • $4$ Newton steps to $ϵ^{16}$ or


                • $3$ two-step cycles to $ϵ^{27}$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 17 '18 at 16:43

























                answered Dec 16 '18 at 21:21









                LutzLLutzL

                59.4k42057




                59.4k42057






























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