Power series, how to shift the index of a summation with a powers 2n+1












1












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I am taking an ordinary differential equation class, and we are currently learning about power series. One thing that comes up is index shifting, for the most part, I can shift the index quite easily, but in the following case, I end up with a fraction index. I have an ODE of the form y'' + 2xy' + y = 0, with one of the solutions being (in power series form):



y = $sum_{n=0}^{infty} b_n x^{2n+1}$



finding y' and y'', gives me:



y' = $sum_{n=1}^{infty} (2n+1)b_n x^{2n}$



y'' = $sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1}$



plugging it in the ODE, I get:



$sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1} + 2xsum_{n=1}^{infty} (2n+1)b_n x^{2n} + sum_{n=0}^{infty} b_n x^{2n+1} = 0$



which is simplified as
$sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1} + sum_{n=1}^{infty} 2(2n+1)b_n x^{2n+1} + sum_{n=0}^{infty} b_n x^{2n+1} = 0$



From there, I can put y and y' together as such:

$b_0x + sum_{n=1}^{infty} [b_n + 2(2n+1)b_n]x^{2n+1} + sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1} = 0$



And now I'm stuck on how to shift the index so I can add both of the remaining summations together. The problem comes from having a power of 2n instead of just n, i end with indexes that are fractions, which cannot be right.



I tried to look around on google but every examples are showing how to shift indexes with a power series of $x^{n+a}$ and not $x^{2n+a}$



Can anyone be kind enough to show me how to shift the index for this problem (and perhaps give me another example which is similar to this)? Thank you






summation differential







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    1












    $begingroup$


    I am taking an ordinary differential equation class, and we are currently learning about power series. One thing that comes up is index shifting, for the most part, I can shift the index quite easily, but in the following case, I end up with a fraction index. I have an ODE of the form y'' + 2xy' + y = 0, with one of the solutions being (in power series form):



    y = $sum_{n=0}^{infty} b_n x^{2n+1}$



    finding y' and y'', gives me:



    y' = $sum_{n=1}^{infty} (2n+1)b_n x^{2n}$



    y'' = $sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1}$



    plugging it in the ODE, I get:



    $sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1} + 2xsum_{n=1}^{infty} (2n+1)b_n x^{2n} + sum_{n=0}^{infty} b_n x^{2n+1} = 0$



    which is simplified as
    $sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1} + sum_{n=1}^{infty} 2(2n+1)b_n x^{2n+1} + sum_{n=0}^{infty} b_n x^{2n+1} = 0$



    From there, I can put y and y' together as such:

    $b_0x + sum_{n=1}^{infty} [b_n + 2(2n+1)b_n]x^{2n+1} + sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1} = 0$



    And now I'm stuck on how to shift the index so I can add both of the remaining summations together. The problem comes from having a power of 2n instead of just n, i end with indexes that are fractions, which cannot be right.



    I tried to look around on google but every examples are showing how to shift indexes with a power series of $x^{n+a}$ and not $x^{2n+a}$



    Can anyone be kind enough to show me how to shift the index for this problem (and perhaps give me another example which is similar to this)? Thank you






    summation differential







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am taking an ordinary differential equation class, and we are currently learning about power series. One thing that comes up is index shifting, for the most part, I can shift the index quite easily, but in the following case, I end up with a fraction index. I have an ODE of the form y'' + 2xy' + y = 0, with one of the solutions being (in power series form):



      y = $sum_{n=0}^{infty} b_n x^{2n+1}$



      finding y' and y'', gives me:



      y' = $sum_{n=1}^{infty} (2n+1)b_n x^{2n}$



      y'' = $sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1}$



      plugging it in the ODE, I get:



      $sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1} + 2xsum_{n=1}^{infty} (2n+1)b_n x^{2n} + sum_{n=0}^{infty} b_n x^{2n+1} = 0$



      which is simplified as
      $sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1} + sum_{n=1}^{infty} 2(2n+1)b_n x^{2n+1} + sum_{n=0}^{infty} b_n x^{2n+1} = 0$



      From there, I can put y and y' together as such:

      $b_0x + sum_{n=1}^{infty} [b_n + 2(2n+1)b_n]x^{2n+1} + sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1} = 0$



      And now I'm stuck on how to shift the index so I can add both of the remaining summations together. The problem comes from having a power of 2n instead of just n, i end with indexes that are fractions, which cannot be right.



      I tried to look around on google but every examples are showing how to shift indexes with a power series of $x^{n+a}$ and not $x^{2n+a}$



      Can anyone be kind enough to show me how to shift the index for this problem (and perhaps give me another example which is similar to this)? Thank you






      summation differential







      share|cite|improve this question









      $endgroup$




      I am taking an ordinary differential equation class, and we are currently learning about power series. One thing that comes up is index shifting, for the most part, I can shift the index quite easily, but in the following case, I end up with a fraction index. I have an ODE of the form y'' + 2xy' + y = 0, with one of the solutions being (in power series form):



      y = $sum_{n=0}^{infty} b_n x^{2n+1}$



      finding y' and y'', gives me:



      y' = $sum_{n=1}^{infty} (2n+1)b_n x^{2n}$



      y'' = $sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1}$



      plugging it in the ODE, I get:



      $sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1} + 2xsum_{n=1}^{infty} (2n+1)b_n x^{2n} + sum_{n=0}^{infty} b_n x^{2n+1} = 0$



      which is simplified as
      $sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1} + sum_{n=1}^{infty} 2(2n+1)b_n x^{2n+1} + sum_{n=0}^{infty} b_n x^{2n+1} = 0$



      From there, I can put y and y' together as such:

      $b_0x + sum_{n=1}^{infty} [b_n + 2(2n+1)b_n]x^{2n+1} + sum_{n=2}^{infty} (2n)(2n+1)b_n x^{2n-1} = 0$



      And now I'm stuck on how to shift the index so I can add both of the remaining summations together. The problem comes from having a power of 2n instead of just n, i end with indexes that are fractions, which cannot be right.



      I tried to look around on google but every examples are showing how to shift indexes with a power series of $x^{n+a}$ and not $x^{2n+a}$



      Can anyone be kind enough to show me how to shift the index for this problem (and perhaps give me another example which is similar to this)? Thank you






      summation differential




      summation differential






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      asked Nov 20 '16 at 23:03









      JohnJohn

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          2 Answers
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          $begingroup$

          Hint



          you can downshift or upshift.



          $$sum_{n=k}^N f(n)=$$



          $$sum_{n=k-1}^{N-1}f(n+1)=sum_{n=k+1}^{N+1}f(n-1)$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            The right-hand sum's lowest term is $x^3$. We want $x^3$ to be of the form $x^{2n+1}$ so that we can combine the right-hand sum with the left-hand one. Clearly then, we can start the sum at n=1 and replace n with n+1 everywhere it appears in the right-hand sum. This sends 2n to 2(n+1), 2n+1 to 2n+3, and $b_n$ to $b_{n+1}$.






            share|cite|improve this answer









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              2 Answers
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              active

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Hint



              you can downshift or upshift.



              $$sum_{n=k}^N f(n)=$$



              $$sum_{n=k-1}^{N-1}f(n+1)=sum_{n=k+1}^{N+1}f(n-1)$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Hint



                you can downshift or upshift.



                $$sum_{n=k}^N f(n)=$$



                $$sum_{n=k-1}^{N-1}f(n+1)=sum_{n=k+1}^{N+1}f(n-1)$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Hint



                  you can downshift or upshift.



                  $$sum_{n=k}^N f(n)=$$



                  $$sum_{n=k-1}^{N-1}f(n+1)=sum_{n=k+1}^{N+1}f(n-1)$$






                  share|cite|improve this answer









                  $endgroup$



                  Hint



                  you can downshift or upshift.



                  $$sum_{n=k}^N f(n)=$$



                  $$sum_{n=k-1}^{N-1}f(n+1)=sum_{n=k+1}^{N+1}f(n-1)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 20 '16 at 23:11









                  hamam_Abdallahhamam_Abdallah

                  38k21634




                  38k21634























                      1












                      $begingroup$

                      The right-hand sum's lowest term is $x^3$. We want $x^3$ to be of the form $x^{2n+1}$ so that we can combine the right-hand sum with the left-hand one. Clearly then, we can start the sum at n=1 and replace n with n+1 everywhere it appears in the right-hand sum. This sends 2n to 2(n+1), 2n+1 to 2n+3, and $b_n$ to $b_{n+1}$.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        The right-hand sum's lowest term is $x^3$. We want $x^3$ to be of the form $x^{2n+1}$ so that we can combine the right-hand sum with the left-hand one. Clearly then, we can start the sum at n=1 and replace n with n+1 everywhere it appears in the right-hand sum. This sends 2n to 2(n+1), 2n+1 to 2n+3, and $b_n$ to $b_{n+1}$.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          The right-hand sum's lowest term is $x^3$. We want $x^3$ to be of the form $x^{2n+1}$ so that we can combine the right-hand sum with the left-hand one. Clearly then, we can start the sum at n=1 and replace n with n+1 everywhere it appears in the right-hand sum. This sends 2n to 2(n+1), 2n+1 to 2n+3, and $b_n$ to $b_{n+1}$.






                          share|cite|improve this answer









                          $endgroup$



                          The right-hand sum's lowest term is $x^3$. We want $x^3$ to be of the form $x^{2n+1}$ so that we can combine the right-hand sum with the left-hand one. Clearly then, we can start the sum at n=1 and replace n with n+1 everywhere it appears in the right-hand sum. This sends 2n to 2(n+1), 2n+1 to 2n+3, and $b_n$ to $b_{n+1}$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 9 '18 at 11:10









                          BelowAverageIntelligenceBelowAverageIntelligence

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