B splines recursion











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Given that by definition the i-th B-spline of order k is:
$$B_{i,k}=w_{i,k}B_{i,k-1}+(1-w_{i+1,k})B_{i+1,k-1}$$



where $w_{j,k}=frac{x-t_j}{t_{j+k-1}-t_j}$



We can define the spline space as
$$S_{k,t}:={sum_ialpha_iB_{i,k}:alpha_i in mathbb{R}} $$



The author pointed out that $$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1}$$
but honestly I don't really get how to arrive to this expression, using the recursion I am able to obtain:
$$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1})$$ which is nothing like the desired result.



Any hint will be greatly appreciated.






functional-analysis recursion spline







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    up vote
    0
    down vote

    favorite












    Given that by definition the i-th B-spline of order k is:
    $$B_{i,k}=w_{i,k}B_{i,k-1}+(1-w_{i+1,k})B_{i+1,k-1}$$



    where $w_{j,k}=frac{x-t_j}{t_{j+k-1}-t_j}$



    We can define the spline space as
    $$S_{k,t}:={sum_ialpha_iB_{i,k}:alpha_i in mathbb{R}} $$



    The author pointed out that $$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1}$$
    but honestly I don't really get how to arrive to this expression, using the recursion I am able to obtain:
    $$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1})$$ which is nothing like the desired result.



    Any hint will be greatly appreciated.






    functional-analysis recursion spline







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Given that by definition the i-th B-spline of order k is:
      $$B_{i,k}=w_{i,k}B_{i,k-1}+(1-w_{i+1,k})B_{i+1,k-1}$$



      where $w_{j,k}=frac{x-t_j}{t_{j+k-1}-t_j}$



      We can define the spline space as
      $$S_{k,t}:={sum_ialpha_iB_{i,k}:alpha_i in mathbb{R}} $$



      The author pointed out that $$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1}$$
      but honestly I don't really get how to arrive to this expression, using the recursion I am able to obtain:
      $$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1})$$ which is nothing like the desired result.



      Any hint will be greatly appreciated.






      functional-analysis recursion spline







      share|cite|improve this question













      Given that by definition the i-th B-spline of order k is:
      $$B_{i,k}=w_{i,k}B_{i,k-1}+(1-w_{i+1,k})B_{i+1,k-1}$$



      where $w_{j,k}=frac{x-t_j}{t_{j+k-1}-t_j}$



      We can define the spline space as
      $$S_{k,t}:={sum_ialpha_iB_{i,k}:alpha_i in mathbb{R}} $$



      The author pointed out that $$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1}$$
      but honestly I don't really get how to arrive to this expression, using the recursion I am able to obtain:
      $$sum_ialpha_i B_{i,k}=sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1})$$ which is nothing like the desired result.



      Any hint will be greatly appreciated.






      functional-analysis recursion spline




      functional-analysis recursion spline






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      share|cite|improve this question











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      asked 2 days ago









      Ramiro Scorolli

      57412




      57412






















          1 Answer
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          The two expressions coincide:
          $$
          begin{split}
          &phantom{=} sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}) \
          &= sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i-1}(1-w_{i,k})B_{i,k-1}) \
          &= sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1} .
          end{split}
          $$






          share|cite|improve this answer





















          • but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
            – Ramiro Scorolli
            yesterday








          • 1




            It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
            – Federico
            yesterday






          • 1




            In both series the same terms are appearing, just shifted.
            – Federico
            yesterday










          • You are right, don't know why I didn't see that. Thanks
            – Ramiro Scorolli
            yesterday











          Your Answer





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          1 Answer
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          The two expressions coincide:
          $$
          begin{split}
          &phantom{=} sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}) \
          &= sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i-1}(1-w_{i,k})B_{i,k-1}) \
          &= sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1} .
          end{split}
          $$






          share|cite|improve this answer





















          • but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
            – Ramiro Scorolli
            yesterday








          • 1




            It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
            – Federico
            yesterday






          • 1




            In both series the same terms are appearing, just shifted.
            – Federico
            yesterday










          • You are right, don't know why I didn't see that. Thanks
            – Ramiro Scorolli
            yesterday















          up vote
          1
          down vote



          accepted










          The two expressions coincide:
          $$
          begin{split}
          &phantom{=} sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}) \
          &= sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i-1}(1-w_{i,k})B_{i,k-1}) \
          &= sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1} .
          end{split}
          $$






          share|cite|improve this answer





















          • but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
            – Ramiro Scorolli
            yesterday








          • 1




            It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
            – Federico
            yesterday






          • 1




            In both series the same terms are appearing, just shifted.
            – Federico
            yesterday










          • You are right, don't know why I didn't see that. Thanks
            – Ramiro Scorolli
            yesterday













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The two expressions coincide:
          $$
          begin{split}
          &phantom{=} sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}) \
          &= sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i-1}(1-w_{i,k})B_{i,k-1}) \
          &= sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1} .
          end{split}
          $$






          share|cite|improve this answer












          The two expressions coincide:
          $$
          begin{split}
          &phantom{=} sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}) \
          &= sum_i(alpha_i w_{i,k}B_{i,k-1}+alpha_{i-1}(1-w_{i,k})B_{i,k-1}) \
          &= sum_i(alpha_i w_{i,k}+alpha_{i-1}(1-w_{i,k}))B_{i,k-1} .
          end{split}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Federico

          2,04658




          2,04658












          • but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
            – Ramiro Scorolli
            yesterday








          • 1




            It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
            – Federico
            yesterday






          • 1




            In both series the same terms are appearing, just shifted.
            – Federico
            yesterday










          • You are right, don't know why I didn't see that. Thanks
            – Ramiro Scorolli
            yesterday


















          • but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
            – Ramiro Scorolli
            yesterday








          • 1




            It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
            – Federico
            yesterday






          • 1




            In both series the same terms are appearing, just shifted.
            – Federico
            yesterday










          • You are right, don't know why I didn't see that. Thanks
            – Ramiro Scorolli
            yesterday
















          but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
          – Ramiro Scorolli
          yesterday






          but why is $alpha_{i}(1-w_{i+1,k})B_{i+1,k-1}=alpha_{i-1}(1-w_{i,k})B_{i,k-1}$?
          – Ramiro Scorolli
          yesterday






          1




          1




          It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
          – Federico
          yesterday




          It is not. The equality is achieved when summing over all possible $i$. For instance $sum_{iinmathbb Z}c_i=sum_{iinmathbb Z}c_{i-1}$.
          – Federico
          yesterday




          1




          1




          In both series the same terms are appearing, just shifted.
          – Federico
          yesterday




          In both series the same terms are appearing, just shifted.
          – Federico
          yesterday












          You are right, don't know why I didn't see that. Thanks
          – Ramiro Scorolli
          yesterday




          You are right, don't know why I didn't see that. Thanks
          – Ramiro Scorolli
          yesterday


















           

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