Find sum $u_0 u_1 + u_1u_2+…+u_{n-2}u_{n-1} $












3












$begingroup$


I have
$$ u_k = cosfrac{2kpi}{n} + i sinfrac{2kpi}{n}$$
And I should calculate:
$$ u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1}+u_{n-1}u_0 $$
But I have stucked:


Firstly I calculate
$$u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1} $$
and put
$$ alpha_k = u_k cdot u_{k+1} = ... = e^{frac{ipi(2k+1)}{n}} $$
and sum
$$ alpha_0 + ... + alpha_{n-2} = ... = e^{frac{ipi}{n}} cdot frac{1-e^{frac{2(n-1)ipi}{n}}}{1-e^{frac{2ipi}{n}}}$$ and I don't know how ti finish that. If it comes to
$$u_{n-1}u_0 = e^{ipi} = -1 $$






complex-analysis summation







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    3












    $begingroup$


    I have
    $$ u_k = cosfrac{2kpi}{n} + i sinfrac{2kpi}{n}$$
    And I should calculate:
    $$ u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1}+u_{n-1}u_0 $$
    But I have stucked:


    Firstly I calculate
    $$u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1} $$
    and put
    $$ alpha_k = u_k cdot u_{k+1} = ... = e^{frac{ipi(2k+1)}{n}} $$
    and sum
    $$ alpha_0 + ... + alpha_{n-2} = ... = e^{frac{ipi}{n}} cdot frac{1-e^{frac{2(n-1)ipi}{n}}}{1-e^{frac{2ipi}{n}}}$$ and I don't know how ti finish that. If it comes to
    $$u_{n-1}u_0 = e^{ipi} = -1 $$






    complex-analysis summation







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      0



      $begingroup$


      I have
      $$ u_k = cosfrac{2kpi}{n} + i sinfrac{2kpi}{n}$$
      And I should calculate:
      $$ u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1}+u_{n-1}u_0 $$
      But I have stucked:


      Firstly I calculate
      $$u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1} $$
      and put
      $$ alpha_k = u_k cdot u_{k+1} = ... = e^{frac{ipi(2k+1)}{n}} $$
      and sum
      $$ alpha_0 + ... + alpha_{n-2} = ... = e^{frac{ipi}{n}} cdot frac{1-e^{frac{2(n-1)ipi}{n}}}{1-e^{frac{2ipi}{n}}}$$ and I don't know how ti finish that. If it comes to
      $$u_{n-1}u_0 = e^{ipi} = -1 $$






      complex-analysis summation







      share|cite|improve this question









      $endgroup$




      I have
      $$ u_k = cosfrac{2kpi}{n} + i sinfrac{2kpi}{n}$$
      And I should calculate:
      $$ u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1}+u_{n-1}u_0 $$
      But I have stucked:


      Firstly I calculate
      $$u_0 u_1 + u_1u_2+...+u_{n-2}u_{n-1} $$
      and put
      $$ alpha_k = u_k cdot u_{k+1} = ... = e^{frac{ipi(2k+1)}{n}} $$
      and sum
      $$ alpha_0 + ... + alpha_{n-2} = ... = e^{frac{ipi}{n}} cdot frac{1-e^{frac{2(n-1)ipi}{n}}}{1-e^{frac{2ipi}{n}}}$$ and I don't know how ti finish that. If it comes to
      $$u_{n-1}u_0 = e^{ipi} = -1 $$






      complex-analysis summation




      complex-analysis summation






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      asked Feb 1 at 16:07









      VirtualUserVirtualUser

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          $begingroup$

          Note that $$alpha_k = u_{k}cdot u_{k+1} = e^{ifrac{2pi}{n}(2k+1)} = e^{ifrac{2pi}{n}}cdot e^{ifrac{4pi}{n}k}.$$
          Therefore,
          $$sum_{k=0}^{n-2}{alpha_k} = e^{ifrac{2pi}{n}}sum_{k=0}^{n-2}{left(e^{ifrac{4pi}{n}}right)^{k}} = e^{ifrac{2pi}{n}} frac{1-e^{ifrac{4pi(n-1)}{n}}}{1-e^{ifrac{4pi}{n}}} = e^{ifrac{2pi}{n}}frac{1-e^{-ifrac{4pi}{n}}}{1-e^{ifrac{4pi}{n}}} = frac{e^{i2pi/n}-e^{-i2pi/n}}{e^{i2pi/n}(e^{-i2pi/n}-e^{i2pi/n})}=-e^{-ifrac{2pi}{n}}.$$






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          $endgroup$





















            1












            $begingroup$

            Hint: The points $u_k$, and therefore $alpha_k$, are distributed completely evenly along the unit circle.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              Here's a way without geometric sums.



              Note that $u_{i+1} = u_i e^{2ipi/n}$ where $u_n:=u_0=1$, thus $sum_{k=0}^{n-1} u_iu_{i+1} = e^{2ipi/n} sum_{k=0}^{n-1} u_i^2$.



              By Newton's identities $ sum_{k=0}^{n-1} u_i^2 = left(sum_{k=0}^{n-1} u_iright)^2- sum_{i,j}u_iu_j$. Since the $u_i$ are the roots of $X^n-1$, $sum_{k=0}^{n-1} u_i=0$ and $sum_{i,j}u_iu_j = 0$, thus $sum_{k=0}^{n-1} u_i^2=0$.






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                Your Answer





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                3 Answers
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                3 Answers
                3






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                $begingroup$

                Note that $$alpha_k = u_{k}cdot u_{k+1} = e^{ifrac{2pi}{n}(2k+1)} = e^{ifrac{2pi}{n}}cdot e^{ifrac{4pi}{n}k}.$$
                Therefore,
                $$sum_{k=0}^{n-2}{alpha_k} = e^{ifrac{2pi}{n}}sum_{k=0}^{n-2}{left(e^{ifrac{4pi}{n}}right)^{k}} = e^{ifrac{2pi}{n}} frac{1-e^{ifrac{4pi(n-1)}{n}}}{1-e^{ifrac{4pi}{n}}} = e^{ifrac{2pi}{n}}frac{1-e^{-ifrac{4pi}{n}}}{1-e^{ifrac{4pi}{n}}} = frac{e^{i2pi/n}-e^{-i2pi/n}}{e^{i2pi/n}(e^{-i2pi/n}-e^{i2pi/n})}=-e^{-ifrac{2pi}{n}}.$$






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                $endgroup$


















                  1












                  $begingroup$

                  Note that $$alpha_k = u_{k}cdot u_{k+1} = e^{ifrac{2pi}{n}(2k+1)} = e^{ifrac{2pi}{n}}cdot e^{ifrac{4pi}{n}k}.$$
                  Therefore,
                  $$sum_{k=0}^{n-2}{alpha_k} = e^{ifrac{2pi}{n}}sum_{k=0}^{n-2}{left(e^{ifrac{4pi}{n}}right)^{k}} = e^{ifrac{2pi}{n}} frac{1-e^{ifrac{4pi(n-1)}{n}}}{1-e^{ifrac{4pi}{n}}} = e^{ifrac{2pi}{n}}frac{1-e^{-ifrac{4pi}{n}}}{1-e^{ifrac{4pi}{n}}} = frac{e^{i2pi/n}-e^{-i2pi/n}}{e^{i2pi/n}(e^{-i2pi/n}-e^{i2pi/n})}=-e^{-ifrac{2pi}{n}}.$$






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Note that $$alpha_k = u_{k}cdot u_{k+1} = e^{ifrac{2pi}{n}(2k+1)} = e^{ifrac{2pi}{n}}cdot e^{ifrac{4pi}{n}k}.$$
                    Therefore,
                    $$sum_{k=0}^{n-2}{alpha_k} = e^{ifrac{2pi}{n}}sum_{k=0}^{n-2}{left(e^{ifrac{4pi}{n}}right)^{k}} = e^{ifrac{2pi}{n}} frac{1-e^{ifrac{4pi(n-1)}{n}}}{1-e^{ifrac{4pi}{n}}} = e^{ifrac{2pi}{n}}frac{1-e^{-ifrac{4pi}{n}}}{1-e^{ifrac{4pi}{n}}} = frac{e^{i2pi/n}-e^{-i2pi/n}}{e^{i2pi/n}(e^{-i2pi/n}-e^{i2pi/n})}=-e^{-ifrac{2pi}{n}}.$$






                    share|cite|improve this answer











                    $endgroup$



                    Note that $$alpha_k = u_{k}cdot u_{k+1} = e^{ifrac{2pi}{n}(2k+1)} = e^{ifrac{2pi}{n}}cdot e^{ifrac{4pi}{n}k}.$$
                    Therefore,
                    $$sum_{k=0}^{n-2}{alpha_k} = e^{ifrac{2pi}{n}}sum_{k=0}^{n-2}{left(e^{ifrac{4pi}{n}}right)^{k}} = e^{ifrac{2pi}{n}} frac{1-e^{ifrac{4pi(n-1)}{n}}}{1-e^{ifrac{4pi}{n}}} = e^{ifrac{2pi}{n}}frac{1-e^{-ifrac{4pi}{n}}}{1-e^{ifrac{4pi}{n}}} = frac{e^{i2pi/n}-e^{-i2pi/n}}{e^{i2pi/n}(e^{-i2pi/n}-e^{i2pi/n})}=-e^{-ifrac{2pi}{n}}.$$







                    share|cite|improve this answer














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                    edited Feb 1 at 18:06

























                    answered Feb 1 at 17:12









                    Math LoverMath Lover

                    14.1k31437




                    14.1k31437























                        1












                        $begingroup$

                        Hint: The points $u_k$, and therefore $alpha_k$, are distributed completely evenly along the unit circle.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Hint: The points $u_k$, and therefore $alpha_k$, are distributed completely evenly along the unit circle.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Hint: The points $u_k$, and therefore $alpha_k$, are distributed completely evenly along the unit circle.






                            share|cite|improve this answer









                            $endgroup$



                            Hint: The points $u_k$, and therefore $alpha_k$, are distributed completely evenly along the unit circle.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 1 at 16:14









                            ArthurArthur

                            122k7122211




                            122k7122211























                                1












                                $begingroup$

                                Here's a way without geometric sums.



                                Note that $u_{i+1} = u_i e^{2ipi/n}$ where $u_n:=u_0=1$, thus $sum_{k=0}^{n-1} u_iu_{i+1} = e^{2ipi/n} sum_{k=0}^{n-1} u_i^2$.



                                By Newton's identities $ sum_{k=0}^{n-1} u_i^2 = left(sum_{k=0}^{n-1} u_iright)^2- sum_{i,j}u_iu_j$. Since the $u_i$ are the roots of $X^n-1$, $sum_{k=0}^{n-1} u_i=0$ and $sum_{i,j}u_iu_j = 0$, thus $sum_{k=0}^{n-1} u_i^2=0$.






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  Here's a way without geometric sums.



                                  Note that $u_{i+1} = u_i e^{2ipi/n}$ where $u_n:=u_0=1$, thus $sum_{k=0}^{n-1} u_iu_{i+1} = e^{2ipi/n} sum_{k=0}^{n-1} u_i^2$.



                                  By Newton's identities $ sum_{k=0}^{n-1} u_i^2 = left(sum_{k=0}^{n-1} u_iright)^2- sum_{i,j}u_iu_j$. Since the $u_i$ are the roots of $X^n-1$, $sum_{k=0}^{n-1} u_i=0$ and $sum_{i,j}u_iu_j = 0$, thus $sum_{k=0}^{n-1} u_i^2=0$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Here's a way without geometric sums.



                                    Note that $u_{i+1} = u_i e^{2ipi/n}$ where $u_n:=u_0=1$, thus $sum_{k=0}^{n-1} u_iu_{i+1} = e^{2ipi/n} sum_{k=0}^{n-1} u_i^2$.



                                    By Newton's identities $ sum_{k=0}^{n-1} u_i^2 = left(sum_{k=0}^{n-1} u_iright)^2- sum_{i,j}u_iu_j$. Since the $u_i$ are the roots of $X^n-1$, $sum_{k=0}^{n-1} u_i=0$ and $sum_{i,j}u_iu_j = 0$, thus $sum_{k=0}^{n-1} u_i^2=0$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Here's a way without geometric sums.



                                    Note that $u_{i+1} = u_i e^{2ipi/n}$ where $u_n:=u_0=1$, thus $sum_{k=0}^{n-1} u_iu_{i+1} = e^{2ipi/n} sum_{k=0}^{n-1} u_i^2$.



                                    By Newton's identities $ sum_{k=0}^{n-1} u_i^2 = left(sum_{k=0}^{n-1} u_iright)^2- sum_{i,j}u_iu_j$. Since the $u_i$ are the roots of $X^n-1$, $sum_{k=0}^{n-1} u_i=0$ and $sum_{i,j}u_iu_j = 0$, thus $sum_{k=0}^{n-1} u_i^2=0$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Feb 1 at 16:31









                                    Gabriel RomonGabriel Romon

                                    18k53387




                                    18k53387






























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