Trouble with changing indexing in summation












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I have trouble with the following: $$P(t) = sum_{i=0}^{n-r}c_i^r(tB^{n-r-1}_{i-1}(t)+(1-t)B_i^{n-r-1}(t))$$



$$= sum_{i=0}^{n-r-1}(tc^r_{i+1}+(1-t)c_i^r)B_i^{n-r-1}(t),$$ where $c_i$ are constants, and $B_i^n(t)$ are the Bernstein polynomials.



I am confused on how the to reach the second equality from the first one. It seems like just a matter of changing the indexing (reducing the sum index from (n-r) to n-(r+1), but I can't decipher how exactly.






summation numerical-methods index-notation







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    1












    $begingroup$


    I have trouble with the following: $$P(t) = sum_{i=0}^{n-r}c_i^r(tB^{n-r-1}_{i-1}(t)+(1-t)B_i^{n-r-1}(t))$$



    $$= sum_{i=0}^{n-r-1}(tc^r_{i+1}+(1-t)c_i^r)B_i^{n-r-1}(t),$$ where $c_i$ are constants, and $B_i^n(t)$ are the Bernstein polynomials.



    I am confused on how the to reach the second equality from the first one. It seems like just a matter of changing the indexing (reducing the sum index from (n-r) to n-(r+1), but I can't decipher how exactly.






    summation numerical-methods index-notation







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I have trouble with the following: $$P(t) = sum_{i=0}^{n-r}c_i^r(tB^{n-r-1}_{i-1}(t)+(1-t)B_i^{n-r-1}(t))$$



      $$= sum_{i=0}^{n-r-1}(tc^r_{i+1}+(1-t)c_i^r)B_i^{n-r-1}(t),$$ where $c_i$ are constants, and $B_i^n(t)$ are the Bernstein polynomials.



      I am confused on how the to reach the second equality from the first one. It seems like just a matter of changing the indexing (reducing the sum index from (n-r) to n-(r+1), but I can't decipher how exactly.






      summation numerical-methods index-notation







      share|cite|improve this question









      $endgroup$




      I have trouble with the following: $$P(t) = sum_{i=0}^{n-r}c_i^r(tB^{n-r-1}_{i-1}(t)+(1-t)B_i^{n-r-1}(t))$$



      $$= sum_{i=0}^{n-r-1}(tc^r_{i+1}+(1-t)c_i^r)B_i^{n-r-1}(t),$$ where $c_i$ are constants, and $B_i^n(t)$ are the Bernstein polynomials.



      I am confused on how the to reach the second equality from the first one. It seems like just a matter of changing the indexing (reducing the sum index from (n-r) to n-(r+1), but I can't decipher how exactly.






      summation numerical-methods index-notation




      summation numerical-methods index-notation






      share|cite|improve this question













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      asked Dec 11 '18 at 9:06









      platypus17platypus17

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          $begingroup$

          Hint:
          begin{align*}
          sum_{i=0}^{n-r}c_i^r&(tB^{n-r-1}_{i-1}(t) +(1-t)B_i^{n-r-1}(t)) \
          &= sum_{i=0}^{n-r}c_i^rtB^{n-r-1}_{i-1}(t)+sum_{i=0}^{n-r}c_i^r(1-t)B_i^{n-r-1}(t) \
          &= sum_{i=-1}^{n-r-1}c_{i+1}^rtB^{n-r-1}_i(t)+sum_{i=0}^{n-r}c_i^r(1-t)B_i^{n-r-1}(t) \
          &= c_0^rtB^{n-r-1}_{-1}(t)+sum_{i=0}^{n-r-1}c_{i+1}^rtB^{n-r-1}_i(t)+sum_{i=0}^{n-r-1}c_i^r(1-t)B_i^{n-r-1}(t)+c_{n-r}^r(1-t)B_{n-r}^{n-r-1}(t)dots.
          end{align*}






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            $begingroup$

            Hint:
            begin{align*}
            sum_{i=0}^{n-r}c_i^r&(tB^{n-r-1}_{i-1}(t) +(1-t)B_i^{n-r-1}(t)) \
            &= sum_{i=0}^{n-r}c_i^rtB^{n-r-1}_{i-1}(t)+sum_{i=0}^{n-r}c_i^r(1-t)B_i^{n-r-1}(t) \
            &= sum_{i=-1}^{n-r-1}c_{i+1}^rtB^{n-r-1}_i(t)+sum_{i=0}^{n-r}c_i^r(1-t)B_i^{n-r-1}(t) \
            &= c_0^rtB^{n-r-1}_{-1}(t)+sum_{i=0}^{n-r-1}c_{i+1}^rtB^{n-r-1}_i(t)+sum_{i=0}^{n-r-1}c_i^r(1-t)B_i^{n-r-1}(t)+c_{n-r}^r(1-t)B_{n-r}^{n-r-1}(t)dots.
            end{align*}






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              $begingroup$

              Hint:
              begin{align*}
              sum_{i=0}^{n-r}c_i^r&(tB^{n-r-1}_{i-1}(t) +(1-t)B_i^{n-r-1}(t)) \
              &= sum_{i=0}^{n-r}c_i^rtB^{n-r-1}_{i-1}(t)+sum_{i=0}^{n-r}c_i^r(1-t)B_i^{n-r-1}(t) \
              &= sum_{i=-1}^{n-r-1}c_{i+1}^rtB^{n-r-1}_i(t)+sum_{i=0}^{n-r}c_i^r(1-t)B_i^{n-r-1}(t) \
              &= c_0^rtB^{n-r-1}_{-1}(t)+sum_{i=0}^{n-r-1}c_{i+1}^rtB^{n-r-1}_i(t)+sum_{i=0}^{n-r-1}c_i^r(1-t)B_i^{n-r-1}(t)+c_{n-r}^r(1-t)B_{n-r}^{n-r-1}(t)dots.
              end{align*}






              share|cite|improve this answer









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                $begingroup$

                Hint:
                begin{align*}
                sum_{i=0}^{n-r}c_i^r&(tB^{n-r-1}_{i-1}(t) +(1-t)B_i^{n-r-1}(t)) \
                &= sum_{i=0}^{n-r}c_i^rtB^{n-r-1}_{i-1}(t)+sum_{i=0}^{n-r}c_i^r(1-t)B_i^{n-r-1}(t) \
                &= sum_{i=-1}^{n-r-1}c_{i+1}^rtB^{n-r-1}_i(t)+sum_{i=0}^{n-r}c_i^r(1-t)B_i^{n-r-1}(t) \
                &= c_0^rtB^{n-r-1}_{-1}(t)+sum_{i=0}^{n-r-1}c_{i+1}^rtB^{n-r-1}_i(t)+sum_{i=0}^{n-r-1}c_i^r(1-t)B_i^{n-r-1}(t)+c_{n-r}^r(1-t)B_{n-r}^{n-r-1}(t)dots.
                end{align*}






                share|cite|improve this answer









                $endgroup$



                Hint:
                begin{align*}
                sum_{i=0}^{n-r}c_i^r&(tB^{n-r-1}_{i-1}(t) +(1-t)B_i^{n-r-1}(t)) \
                &= sum_{i=0}^{n-r}c_i^rtB^{n-r-1}_{i-1}(t)+sum_{i=0}^{n-r}c_i^r(1-t)B_i^{n-r-1}(t) \
                &= sum_{i=-1}^{n-r-1}c_{i+1}^rtB^{n-r-1}_i(t)+sum_{i=0}^{n-r}c_i^r(1-t)B_i^{n-r-1}(t) \
                &= c_0^rtB^{n-r-1}_{-1}(t)+sum_{i=0}^{n-r-1}c_{i+1}^rtB^{n-r-1}_i(t)+sum_{i=0}^{n-r-1}c_i^r(1-t)B_i^{n-r-1}(t)+c_{n-r}^r(1-t)B_{n-r}^{n-r-1}(t)dots.
                end{align*}







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                answered Dec 11 '18 at 9:28









                Greg MartinGreg Martin

                35.3k23263




                35.3k23263






























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