Scaling forcing function before numerical integration of linear ODE












1












$begingroup$


Suppose that we have a linear ode which we want to integrate numerically:
$$
L{y}=f(t)
$$
where $L$ is a linear differential operator (and some initial conditions), by using a method of order $mathcal{O}(h^n)$ -- $h$ being the step size of the method.



Is it true that if we scale up the forcing function $f(t)$ with a factor $K$ before integrating, and then rescale down the solution $y$ by $frac{1}{K}$ we have increased the precision of the method? The argument being that the order of the method is still the same and only dependent on the step size. So when we scale down the error is reduced.






ordinary-differential-equations numerical-methods soft-question







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Suppose that we have a linear ode which we want to integrate numerically:
    $$
    L{y}=f(t)
    $$
    where $L$ is a linear differential operator (and some initial conditions), by using a method of order $mathcal{O}(h^n)$ -- $h$ being the step size of the method.



    Is it true that if we scale up the forcing function $f(t)$ with a factor $K$ before integrating, and then rescale down the solution $y$ by $frac{1}{K}$ we have increased the precision of the method? The argument being that the order of the method is still the same and only dependent on the step size. So when we scale down the error is reduced.






    ordinary-differential-equations numerical-methods soft-question







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Suppose that we have a linear ode which we want to integrate numerically:
      $$
      L{y}=f(t)
      $$
      where $L$ is a linear differential operator (and some initial conditions), by using a method of order $mathcal{O}(h^n)$ -- $h$ being the step size of the method.



      Is it true that if we scale up the forcing function $f(t)$ with a factor $K$ before integrating, and then rescale down the solution $y$ by $frac{1}{K}$ we have increased the precision of the method? The argument being that the order of the method is still the same and only dependent on the step size. So when we scale down the error is reduced.






      ordinary-differential-equations numerical-methods soft-question







      share|cite|improve this question









      $endgroup$




      Suppose that we have a linear ode which we want to integrate numerically:
      $$
      L{y}=f(t)
      $$
      where $L$ is a linear differential operator (and some initial conditions), by using a method of order $mathcal{O}(h^n)$ -- $h$ being the step size of the method.



      Is it true that if we scale up the forcing function $f(t)$ with a factor $K$ before integrating, and then rescale down the solution $y$ by $frac{1}{K}$ we have increased the precision of the method? The argument being that the order of the method is still the same and only dependent on the step size. So when we scale down the error is reduced.






      ordinary-differential-equations numerical-methods soft-question




      ordinary-differential-equations numerical-methods soft-question






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 20:31









      AmbeshAmbesh

      1,4201039




      1,4201039






















          1 Answer
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          active

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          3












          $begingroup$

          Imagine the function $y_s$ is the actual solution to the problem



          $$
          L{y_s} = f(t) tag{1a}
          $$



          And $hat{y}$ is an approximation, the error is measured as



          $$
          epsilon(n) = |y_s(t_n) - hat{y}_n| tag{2a}
          $$



          Now, since $L$ is linear, if you multiply both sides of (1a) by $K$, then



          $$
          L{ Ky_s } = K f(t) tag{1b}
          $$



          And the approximated solution is $hat{y}^{K}$ so the error is



          $$
          epsilon^K(n) = |K y_s(t_n) - hat{y}^{K}| = Kleft|y_s(t_n) - frac{hat{y}^{K}}{K}right| = K epsilon(n) tag{2b}
          $$



          So the error also scales as $K$. That is: if you multiply by $K$ and then divide the result, the accuracy does not increase






          share|cite|improve this answer









          $endgroup$













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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Imagine the function $y_s$ is the actual solution to the problem



            $$
            L{y_s} = f(t) tag{1a}
            $$



            And $hat{y}$ is an approximation, the error is measured as



            $$
            epsilon(n) = |y_s(t_n) - hat{y}_n| tag{2a}
            $$



            Now, since $L$ is linear, if you multiply both sides of (1a) by $K$, then



            $$
            L{ Ky_s } = K f(t) tag{1b}
            $$



            And the approximated solution is $hat{y}^{K}$ so the error is



            $$
            epsilon^K(n) = |K y_s(t_n) - hat{y}^{K}| = Kleft|y_s(t_n) - frac{hat{y}^{K}}{K}right| = K epsilon(n) tag{2b}
            $$



            So the error also scales as $K$. That is: if you multiply by $K$ and then divide the result, the accuracy does not increase






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Imagine the function $y_s$ is the actual solution to the problem



              $$
              L{y_s} = f(t) tag{1a}
              $$



              And $hat{y}$ is an approximation, the error is measured as



              $$
              epsilon(n) = |y_s(t_n) - hat{y}_n| tag{2a}
              $$



              Now, since $L$ is linear, if you multiply both sides of (1a) by $K$, then



              $$
              L{ Ky_s } = K f(t) tag{1b}
              $$



              And the approximated solution is $hat{y}^{K}$ so the error is



              $$
              epsilon^K(n) = |K y_s(t_n) - hat{y}^{K}| = Kleft|y_s(t_n) - frac{hat{y}^{K}}{K}right| = K epsilon(n) tag{2b}
              $$



              So the error also scales as $K$. That is: if you multiply by $K$ and then divide the result, the accuracy does not increase






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Imagine the function $y_s$ is the actual solution to the problem



                $$
                L{y_s} = f(t) tag{1a}
                $$



                And $hat{y}$ is an approximation, the error is measured as



                $$
                epsilon(n) = |y_s(t_n) - hat{y}_n| tag{2a}
                $$



                Now, since $L$ is linear, if you multiply both sides of (1a) by $K$, then



                $$
                L{ Ky_s } = K f(t) tag{1b}
                $$



                And the approximated solution is $hat{y}^{K}$ so the error is



                $$
                epsilon^K(n) = |K y_s(t_n) - hat{y}^{K}| = Kleft|y_s(t_n) - frac{hat{y}^{K}}{K}right| = K epsilon(n) tag{2b}
                $$



                So the error also scales as $K$. That is: if you multiply by $K$ and then divide the result, the accuracy does not increase






                share|cite|improve this answer









                $endgroup$



                Imagine the function $y_s$ is the actual solution to the problem



                $$
                L{y_s} = f(t) tag{1a}
                $$



                And $hat{y}$ is an approximation, the error is measured as



                $$
                epsilon(n) = |y_s(t_n) - hat{y}_n| tag{2a}
                $$



                Now, since $L$ is linear, if you multiply both sides of (1a) by $K$, then



                $$
                L{ Ky_s } = K f(t) tag{1b}
                $$



                And the approximated solution is $hat{y}^{K}$ so the error is



                $$
                epsilon^K(n) = |K y_s(t_n) - hat{y}^{K}| = Kleft|y_s(t_n) - frac{hat{y}^{K}}{K}right| = K epsilon(n) tag{2b}
                $$



                So the error also scales as $K$. That is: if you multiply by $K$ and then divide the result, the accuracy does not increase







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 7 at 21:07









                caveraccaverac

                14.5k31130




                14.5k31130






























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