Recurrence relation (numerical analysis/errors)












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Let $p_{k+1}=a_kp_k+b_kp_{k-1}, k in mathbb{N}$ be a recurrence relation with coefficients $a_k, b_k in mathbb{R}$ and start values $p_0, p_1 in mathbb{R}$.



So the values of starting and the coefficients are bugged:



$tilde{p}_0=p_0(1+delta_0), tilde{p}_1=p_1(1+delta_1), tilde{a}_k=a_k(1+alpha_k), tilde{b}_k=b_k(1+beta_k)$



So bugged values $tilde{p}_k$ emerge.



How to find a recurrence relation for the absolute and relative error:



$Delta_{Abs}(p_{k+1}):=tilde{p}_{k+1}-p_{k+1}, Delta_{Rel}(p_{k+1}):=frac{Delta_{abs}(p_{k+1})}{p_{k+1}}$?



I used this for the absolute error:



$Delta(p_{k+1})=a_kDelta p_k+b_kDelta p_{k-1}$



So with the start values $Delta p_0=delta_0 p_0$ and $Delta p_1= delta_1p_1$ I get the recursion:



$alpha_ka_ktilde{p}_{k-1}+beta_kb_ktilde{p}_{k-1}=alpha_ka_kp_k+beta_kb_kp_{k-1}$



Same for the relative error.



Is this recurrence relation correct or has it to be done differently?






proof-verification numerical-methods recurrence-relations







share|cite|improve this question





























    0














    Let $p_{k+1}=a_kp_k+b_kp_{k-1}, k in mathbb{N}$ be a recurrence relation with coefficients $a_k, b_k in mathbb{R}$ and start values $p_0, p_1 in mathbb{R}$.



    So the values of starting and the coefficients are bugged:



    $tilde{p}_0=p_0(1+delta_0), tilde{p}_1=p_1(1+delta_1), tilde{a}_k=a_k(1+alpha_k), tilde{b}_k=b_k(1+beta_k)$



    So bugged values $tilde{p}_k$ emerge.



    How to find a recurrence relation for the absolute and relative error:



    $Delta_{Abs}(p_{k+1}):=tilde{p}_{k+1}-p_{k+1}, Delta_{Rel}(p_{k+1}):=frac{Delta_{abs}(p_{k+1})}{p_{k+1}}$?



    I used this for the absolute error:



    $Delta(p_{k+1})=a_kDelta p_k+b_kDelta p_{k-1}$



    So with the start values $Delta p_0=delta_0 p_0$ and $Delta p_1= delta_1p_1$ I get the recursion:



    $alpha_ka_ktilde{p}_{k-1}+beta_kb_ktilde{p}_{k-1}=alpha_ka_kp_k+beta_kb_kp_{k-1}$



    Same for the relative error.



    Is this recurrence relation correct or has it to be done differently?






    proof-verification numerical-methods recurrence-relations







    share|cite|improve this question



























      0












      0








      0







      Let $p_{k+1}=a_kp_k+b_kp_{k-1}, k in mathbb{N}$ be a recurrence relation with coefficients $a_k, b_k in mathbb{R}$ and start values $p_0, p_1 in mathbb{R}$.



      So the values of starting and the coefficients are bugged:



      $tilde{p}_0=p_0(1+delta_0), tilde{p}_1=p_1(1+delta_1), tilde{a}_k=a_k(1+alpha_k), tilde{b}_k=b_k(1+beta_k)$



      So bugged values $tilde{p}_k$ emerge.



      How to find a recurrence relation for the absolute and relative error:



      $Delta_{Abs}(p_{k+1}):=tilde{p}_{k+1}-p_{k+1}, Delta_{Rel}(p_{k+1}):=frac{Delta_{abs}(p_{k+1})}{p_{k+1}}$?



      I used this for the absolute error:



      $Delta(p_{k+1})=a_kDelta p_k+b_kDelta p_{k-1}$



      So with the start values $Delta p_0=delta_0 p_0$ and $Delta p_1= delta_1p_1$ I get the recursion:



      $alpha_ka_ktilde{p}_{k-1}+beta_kb_ktilde{p}_{k-1}=alpha_ka_kp_k+beta_kb_kp_{k-1}$



      Same for the relative error.



      Is this recurrence relation correct or has it to be done differently?






      proof-verification numerical-methods recurrence-relations







      share|cite|improve this question















      Let $p_{k+1}=a_kp_k+b_kp_{k-1}, k in mathbb{N}$ be a recurrence relation with coefficients $a_k, b_k in mathbb{R}$ and start values $p_0, p_1 in mathbb{R}$.



      So the values of starting and the coefficients are bugged:



      $tilde{p}_0=p_0(1+delta_0), tilde{p}_1=p_1(1+delta_1), tilde{a}_k=a_k(1+alpha_k), tilde{b}_k=b_k(1+beta_k)$



      So bugged values $tilde{p}_k$ emerge.



      How to find a recurrence relation for the absolute and relative error:



      $Delta_{Abs}(p_{k+1}):=tilde{p}_{k+1}-p_{k+1}, Delta_{Rel}(p_{k+1}):=frac{Delta_{abs}(p_{k+1})}{p_{k+1}}$?



      I used this for the absolute error:



      $Delta(p_{k+1})=a_kDelta p_k+b_kDelta p_{k-1}$



      So with the start values $Delta p_0=delta_0 p_0$ and $Delta p_1= delta_1p_1$ I get the recursion:



      $alpha_ka_ktilde{p}_{k-1}+beta_kb_ktilde{p}_{k-1}=alpha_ka_kp_k+beta_kb_kp_{k-1}$



      Same for the relative error.



      Is this recurrence relation correct or has it to be done differently?






      proof-verification numerical-methods recurrence-relations




      proof-verification numerical-methods recurrence-relations






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 24 at 14:57

























      asked Nov 24 at 13:15









      Nekarts

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          You forgot to consider the absolute error of the coefficients:
          $$Delta(p_{k+1})=a_kDelta p_k +b_kDelta p_{k-1}+p_kDelta a_k+p_{k-1}Delta b_k.$$



          The problem with the relative error is the possible cancellation hidden in the addition (while for the single summands, the relative error of $a_kp_k$ is just the sum of the relative errors of the factors). If we knew for example that all numbers involved are positive, we'd be in a much better shape, namely, $max{Delta_{rel}(a_k)+Delta_{rel}(p_k),Delta_{rel}(b_k)+Delta_{rel}(p_{k-1})}$, and this is readily bounded by $(k+1)delta$ (induction on $k$) if all involved numbers have relative error $ledelta$.






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            You forgot to consider the absolute error of the coefficients:
            $$Delta(p_{k+1})=a_kDelta p_k +b_kDelta p_{k-1}+p_kDelta a_k+p_{k-1}Delta b_k.$$



            The problem with the relative error is the possible cancellation hidden in the addition (while for the single summands, the relative error of $a_kp_k$ is just the sum of the relative errors of the factors). If we knew for example that all numbers involved are positive, we'd be in a much better shape, namely, $max{Delta_{rel}(a_k)+Delta_{rel}(p_k),Delta_{rel}(b_k)+Delta_{rel}(p_{k-1})}$, and this is readily bounded by $(k+1)delta$ (induction on $k$) if all involved numbers have relative error $ledelta$.






            share|cite|improve this answer


























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              You forgot to consider the absolute error of the coefficients:
              $$Delta(p_{k+1})=a_kDelta p_k +b_kDelta p_{k-1}+p_kDelta a_k+p_{k-1}Delta b_k.$$



              The problem with the relative error is the possible cancellation hidden in the addition (while for the single summands, the relative error of $a_kp_k$ is just the sum of the relative errors of the factors). If we knew for example that all numbers involved are positive, we'd be in a much better shape, namely, $max{Delta_{rel}(a_k)+Delta_{rel}(p_k),Delta_{rel}(b_k)+Delta_{rel}(p_{k-1})}$, and this is readily bounded by $(k+1)delta$ (induction on $k$) if all involved numbers have relative error $ledelta$.






              share|cite|improve this answer
























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                You forgot to consider the absolute error of the coefficients:
                $$Delta(p_{k+1})=a_kDelta p_k +b_kDelta p_{k-1}+p_kDelta a_k+p_{k-1}Delta b_k.$$



                The problem with the relative error is the possible cancellation hidden in the addition (while for the single summands, the relative error of $a_kp_k$ is just the sum of the relative errors of the factors). If we knew for example that all numbers involved are positive, we'd be in a much better shape, namely, $max{Delta_{rel}(a_k)+Delta_{rel}(p_k),Delta_{rel}(b_k)+Delta_{rel}(p_{k-1})}$, and this is readily bounded by $(k+1)delta$ (induction on $k$) if all involved numbers have relative error $ledelta$.






                share|cite|improve this answer












                You forgot to consider the absolute error of the coefficients:
                $$Delta(p_{k+1})=a_kDelta p_k +b_kDelta p_{k-1}+p_kDelta a_k+p_{k-1}Delta b_k.$$



                The problem with the relative error is the possible cancellation hidden in the addition (while for the single summands, the relative error of $a_kp_k$ is just the sum of the relative errors of the factors). If we knew for example that all numbers involved are positive, we'd be in a much better shape, namely, $max{Delta_{rel}(a_k)+Delta_{rel}(p_k),Delta_{rel}(b_k)+Delta_{rel}(p_{k-1})}$, and this is readily bounded by $(k+1)delta$ (induction on $k$) if all involved numbers have relative error $ledelta$.







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Nov 25 at 11:22









                Hagen von Eitzen

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