Expression for sum of n exponentials












4












$begingroup$


So I have this sum of exponentials and I would like to find an expression for it.



$$sum^n_{i=1} e^{mu(i-1)} $$



Note that $i$ is not an imaginary indicator. I am aware there is a formula for summing a purely exponential sum, but I am not clear as to what happens to the $mu$.






sequences-and-series summation geometric-series







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    So I have this sum of exponentials and I would like to find an expression for it.



    $$sum^n_{i=1} e^{mu(i-1)} $$



    Note that $i$ is not an imaginary indicator. I am aware there is a formula for summing a purely exponential sum, but I am not clear as to what happens to the $mu$.






    sequences-and-series summation geometric-series







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      So I have this sum of exponentials and I would like to find an expression for it.



      $$sum^n_{i=1} e^{mu(i-1)} $$



      Note that $i$ is not an imaginary indicator. I am aware there is a formula for summing a purely exponential sum, but I am not clear as to what happens to the $mu$.






      sequences-and-series summation geometric-series







      share|cite|improve this question









      $endgroup$




      So I have this sum of exponentials and I would like to find an expression for it.



      $$sum^n_{i=1} e^{mu(i-1)} $$



      Note that $i$ is not an imaginary indicator. I am aware there is a formula for summing a purely exponential sum, but I am not clear as to what happens to the $mu$.






      sequences-and-series summation geometric-series




      sequences-and-series summation geometric-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 7 hours ago









      DioDio

      928




      928






















          4 Answers
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          6












          $begingroup$

          The sum $S$ can be rewritten as $$S= sum_{i=0}^{n-1} e^{mu i}=sum_{i=0}^{n-1} (e^{mu})^i=frac{1-e^{mu n}}{1-e^{mu}}$$ since the geometric series $$sum_{i=0}^{n-1} x^{i}=frac{1-x^n}{1-x}$$






          share|cite|improve this answer









          $endgroup$





















            5












            $begingroup$

            Define $a=e^mu$ when $mune 0$. Then you have $$sum^n_{i=1} e^{mu(i-1)} =sum^n_{i=1} a^{i-1}=1+a+cdots+a^{n-1}={a^n-1over a-1}={e^{mu n}-1over e^mu -1} $$For $mu =0 $ we obtain$$sum^n_{i=1} e^{mu(i-1)}=n$$






            share|cite|improve this answer









            $endgroup$





















              3












              $begingroup$

              One may recall that
              $$
              sum_{i=1}^nx^{i-1}=frac{1-x^n}{1-x},qquad xneq1.
              $$               What if you put $x=e^mu$?






              share|cite|improve this answer









              $endgroup$





















                3












                $begingroup$

                Hint:



                This the sum of the $n$ first terms of the geometric series with ratio $mathrm e^mu$, since $;mathrm e^{mu(i-1)}=(mathrm e^{mu})^{i-1}$.






                share|cite|improve this answer









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                  4 Answers
                  4






                  active

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                  4 Answers
                  4






                  active

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                  active

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                  active

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                  6












                  $begingroup$

                  The sum $S$ can be rewritten as $$S= sum_{i=0}^{n-1} e^{mu i}=sum_{i=0}^{n-1} (e^{mu})^i=frac{1-e^{mu n}}{1-e^{mu}}$$ since the geometric series $$sum_{i=0}^{n-1} x^{i}=frac{1-x^n}{1-x}$$






                  share|cite|improve this answer









                  $endgroup$


















                    6












                    $begingroup$

                    The sum $S$ can be rewritten as $$S= sum_{i=0}^{n-1} e^{mu i}=sum_{i=0}^{n-1} (e^{mu})^i=frac{1-e^{mu n}}{1-e^{mu}}$$ since the geometric series $$sum_{i=0}^{n-1} x^{i}=frac{1-x^n}{1-x}$$






                    share|cite|improve this answer









                    $endgroup$
















                      6












                      6








                      6





                      $begingroup$

                      The sum $S$ can be rewritten as $$S= sum_{i=0}^{n-1} e^{mu i}=sum_{i=0}^{n-1} (e^{mu})^i=frac{1-e^{mu n}}{1-e^{mu}}$$ since the geometric series $$sum_{i=0}^{n-1} x^{i}=frac{1-x^n}{1-x}$$






                      share|cite|improve this answer









                      $endgroup$



                      The sum $S$ can be rewritten as $$S= sum_{i=0}^{n-1} e^{mu i}=sum_{i=0}^{n-1} (e^{mu})^i=frac{1-e^{mu n}}{1-e^{mu}}$$ since the geometric series $$sum_{i=0}^{n-1} x^{i}=frac{1-x^n}{1-x}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 7 hours ago









                      aledenaleden

                      1,962511




                      1,962511























                          5












                          $begingroup$

                          Define $a=e^mu$ when $mune 0$. Then you have $$sum^n_{i=1} e^{mu(i-1)} =sum^n_{i=1} a^{i-1}=1+a+cdots+a^{n-1}={a^n-1over a-1}={e^{mu n}-1over e^mu -1} $$For $mu =0 $ we obtain$$sum^n_{i=1} e^{mu(i-1)}=n$$






                          share|cite|improve this answer









                          $endgroup$


















                            5












                            $begingroup$

                            Define $a=e^mu$ when $mune 0$. Then you have $$sum^n_{i=1} e^{mu(i-1)} =sum^n_{i=1} a^{i-1}=1+a+cdots+a^{n-1}={a^n-1over a-1}={e^{mu n}-1over e^mu -1} $$For $mu =0 $ we obtain$$sum^n_{i=1} e^{mu(i-1)}=n$$






                            share|cite|improve this answer









                            $endgroup$
















                              5












                              5








                              5





                              $begingroup$

                              Define $a=e^mu$ when $mune 0$. Then you have $$sum^n_{i=1} e^{mu(i-1)} =sum^n_{i=1} a^{i-1}=1+a+cdots+a^{n-1}={a^n-1over a-1}={e^{mu n}-1over e^mu -1} $$For $mu =0 $ we obtain$$sum^n_{i=1} e^{mu(i-1)}=n$$






                              share|cite|improve this answer









                              $endgroup$



                              Define $a=e^mu$ when $mune 0$. Then you have $$sum^n_{i=1} e^{mu(i-1)} =sum^n_{i=1} a^{i-1}=1+a+cdots+a^{n-1}={a^n-1over a-1}={e^{mu n}-1over e^mu -1} $$For $mu =0 $ we obtain$$sum^n_{i=1} e^{mu(i-1)}=n$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 7 hours ago









                              Mostafa AyazMostafa Ayaz

                              14.8k3938




                              14.8k3938























                                  3












                                  $begingroup$

                                  One may recall that
                                  $$
                                  sum_{i=1}^nx^{i-1}=frac{1-x^n}{1-x},qquad xneq1.
                                  $$                                   What if you put $x=e^mu$?






                                  share|cite|improve this answer









                                  $endgroup$


















                                    3












                                    $begingroup$

                                    One may recall that
                                    $$
                                    sum_{i=1}^nx^{i-1}=frac{1-x^n}{1-x},qquad xneq1.
                                    $$                                     What if you put $x=e^mu$?






                                    share|cite|improve this answer









                                    $endgroup$
















                                      3












                                      3








                                      3





                                      $begingroup$

                                      One may recall that
                                      $$
                                      sum_{i=1}^nx^{i-1}=frac{1-x^n}{1-x},qquad xneq1.
                                      $$                                       What if you put $x=e^mu$?






                                      share|cite|improve this answer









                                      $endgroup$



                                      One may recall that
                                      $$
                                      sum_{i=1}^nx^{i-1}=frac{1-x^n}{1-x},qquad xneq1.
                                      $$                                       What if you put $x=e^mu$?







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 7 hours ago









                                      Olivier OloaOlivier Oloa

                                      108k17176293




                                      108k17176293























                                          3












                                          $begingroup$

                                          Hint:



                                          This the sum of the $n$ first terms of the geometric series with ratio $mathrm e^mu$, since $;mathrm e^{mu(i-1)}=(mathrm e^{mu})^{i-1}$.






                                          share|cite|improve this answer









                                          $endgroup$


















                                            3












                                            $begingroup$

                                            Hint:



                                            This the sum of the $n$ first terms of the geometric series with ratio $mathrm e^mu$, since $;mathrm e^{mu(i-1)}=(mathrm e^{mu})^{i-1}$.






                                            share|cite|improve this answer









                                            $endgroup$
















                                              3












                                              3








                                              3





                                              $begingroup$

                                              Hint:



                                              This the sum of the $n$ first terms of the geometric series with ratio $mathrm e^mu$, since $;mathrm e^{mu(i-1)}=(mathrm e^{mu})^{i-1}$.






                                              share|cite|improve this answer









                                              $endgroup$



                                              Hint:



                                              This the sum of the $n$ first terms of the geometric series with ratio $mathrm e^mu$, since $;mathrm e^{mu(i-1)}=(mathrm e^{mu})^{i-1}$.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered 7 hours ago









                                              BernardBernard

                                              119k639112




                                              119k639112






























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