Iterated backward difference quotient from splines

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I am working on a certain problem: say we have a function $f$ defined on an interval $[a,b]$ and we want the spline $Sf$ of order $k$ (and knots needed for the B-splines are in $[a,b]$) to agree to $f$ and its first $k-1$ derivatives at $a$. In doing so, I came across the following:



Suppose we have a system of the form ($j in mathbbZ$ fixed)
beginalign*
f^(m)(a) = sum_i = (j-1)k+1^jk alpha_i^(m+1) B_i,k-m(a), quad m = 0, 1, ldots, k-1,
endalign*
where $B_i,r$ is a B-spline of order $r$ and $alpha_i^(1) = alpha_i$ and $alpha_i^(m+1) = (k-m)cdot dfracalpha_i^(m) - alpha_i-1^(m)t_i+k-m - t_i$ for $m >0$.



Is there a neat way to formulate this into a matrix equation with the $alpha_i$ as the unknowns?







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    I am working on a certain problem: say we have a function $f$ defined on an interval $[a,b]$ and we want the spline $Sf$ of order $k$ (and knots needed for the B-splines are in $[a,b]$) to agree to $f$ and its first $k-1$ derivatives at $a$. In doing so, I came across the following:



    Suppose we have a system of the form ($j in mathbbZ$ fixed)
    beginalign*
    f^(m)(a) = sum_i = (j-1)k+1^jk alpha_i^(m+1) B_i,k-m(a), quad m = 0, 1, ldots, k-1,
    endalign*
    where $B_i,r$ is a B-spline of order $r$ and $alpha_i^(1) = alpha_i$ and $alpha_i^(m+1) = (k-m)cdot dfracalpha_i^(m) - alpha_i-1^(m)t_i+k-m - t_i$ for $m >0$.



    Is there a neat way to formulate this into a matrix equation with the $alpha_i$ as the unknowns?







    share|cite|improve this question






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I am working on a certain problem: say we have a function $f$ defined on an interval $[a,b]$ and we want the spline $Sf$ of order $k$ (and knots needed for the B-splines are in $[a,b]$) to agree to $f$ and its first $k-1$ derivatives at $a$. In doing so, I came across the following:



      Suppose we have a system of the form ($j in mathbbZ$ fixed)
      beginalign*
      f^(m)(a) = sum_i = (j-1)k+1^jk alpha_i^(m+1) B_i,k-m(a), quad m = 0, 1, ldots, k-1,
      endalign*
      where $B_i,r$ is a B-spline of order $r$ and $alpha_i^(1) = alpha_i$ and $alpha_i^(m+1) = (k-m)cdot dfracalpha_i^(m) - alpha_i-1^(m)t_i+k-m - t_i$ for $m >0$.



      Is there a neat way to formulate this into a matrix equation with the $alpha_i$ as the unknowns?







      share|cite|improve this question












      I am working on a certain problem: say we have a function $f$ defined on an interval $[a,b]$ and we want the spline $Sf$ of order $k$ (and knots needed for the B-splines are in $[a,b]$) to agree to $f$ and its first $k-1$ derivatives at $a$. In doing so, I came across the following:



      Suppose we have a system of the form ($j in mathbbZ$ fixed)
      beginalign*
      f^(m)(a) = sum_i = (j-1)k+1^jk alpha_i^(m+1) B_i,k-m(a), quad m = 0, 1, ldots, k-1,
      endalign*
      where $B_i,r$ is a B-spline of order $r$ and $alpha_i^(1) = alpha_i$ and $alpha_i^(m+1) = (k-m)cdot dfracalpha_i^(m) - alpha_i-1^(m)t_i+k-m - t_i$ for $m >0$.



      Is there a neat way to formulate this into a matrix equation with the $alpha_i$ as the unknowns?









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 15 at 7:52









      Erdberg

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