Trouble with changing indexing in summation












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












    $begingroup$


    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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 11 '18 at 9:06









      platypus17platypus17

      296




      296






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          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









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035091%2ftrouble-with-changing-indexing-in-summation%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            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









            $endgroup$


















              1












              $begingroup$

              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









              $endgroup$
















                1












                1








                1





                $begingroup$

                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*}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 11 '18 at 9:28









                Greg MartinGreg Martin

                35.3k23263




                35.3k23263






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035091%2ftrouble-with-changing-indexing-in-summation%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Plaza Victoria

                    Image
                    Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                    Read more

                    In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
                    Read more

                    How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

                    Image
                    1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
                    Read more