Evaluate $sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n}$












1












$begingroup$


I am trying to evaluate
$$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n}$$
where $ninmathbb{Z}$ and $x,zinmathbb{Z}_{2^n}$.



Clearly, if $x=z$, then $sum=2^n$. But I am unsure about when $xneq z$. I believe it to be $=0$ as that would make my problem self-consistent.






algebra-precalculus summation exponential-sum







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  • 1




    $begingroup$
    It's a geometric sequence in $y$.
    $endgroup$
    – Simply Beautiful Art
    Jan 16 at 14:39
















1












$begingroup$


I am trying to evaluate
$$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n}$$
where $ninmathbb{Z}$ and $x,zinmathbb{Z}_{2^n}$.



Clearly, if $x=z$, then $sum=2^n$. But I am unsure about when $xneq z$. I believe it to be $=0$ as that would make my problem self-consistent.






algebra-precalculus summation exponential-sum







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It's a geometric sequence in $y$.
    $endgroup$
    – Simply Beautiful Art
    Jan 16 at 14:39














1












1








1





$begingroup$


I am trying to evaluate
$$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n}$$
where $ninmathbb{Z}$ and $x,zinmathbb{Z}_{2^n}$.



Clearly, if $x=z$, then $sum=2^n$. But I am unsure about when $xneq z$. I believe it to be $=0$ as that would make my problem self-consistent.






algebra-precalculus summation exponential-sum







share|cite|improve this question









$endgroup$




I am trying to evaluate
$$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n}$$
where $ninmathbb{Z}$ and $x,zinmathbb{Z}_{2^n}$.



Clearly, if $x=z$, then $sum=2^n$. But I am unsure about when $xneq z$. I believe it to be $=0$ as that would make my problem self-consistent.






algebra-precalculus summation exponential-sum




algebra-precalculus summation exponential-sum






share|cite|improve this question













share|cite|improve this question











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asked Jan 16 at 14:36









DanDan

447




447








  • 1




    $begingroup$
    It's a geometric sequence in $y$.
    $endgroup$
    – Simply Beautiful Art
    Jan 16 at 14:39














  • 1




    $begingroup$
    It's a geometric sequence in $y$.
    $endgroup$
    – Simply Beautiful Art
    Jan 16 at 14:39








1




1




$begingroup$
It's a geometric sequence in $y$.
$endgroup$
– Simply Beautiful Art
Jan 16 at 14:39




$begingroup$
It's a geometric sequence in $y$.
$endgroup$
– Simply Beautiful Art
Jan 16 at 14:39










2 Answers
2






active

oldest

votes


















2












$begingroup$

We have two cases:




Case $1$: $z neq x$ (mod. $2^n$) $$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n} = sum_{y=0}^{2^n-1} a^y = frac{1 - a^{2^n}}{1-a}$$
where $a = Big( e^{2pi i(z-x)/2^n} Big)$ Replacing as @lab bhattacharjee did, we get $0$.




and




Case $2$: $z = x$ (mod. $2^n$) $$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n} = sum_{y=0}^{2^n-1} e^0 = 2^n$$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    What if $2^n$ divide $z-x$
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 15:15










  • $begingroup$
    that's a remarkable note .. should be added as follows $z neq x $ (mod. $2^n$).
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 15:26



















1












$begingroup$

$$sum_{y=0}^{2^n-1}(e^{2pi i(z-x)/2^n})^y$$



$$=dfrac{(e^{2pi i(z-x)/2^n})^{2^n}-1}{e^{2pi i(z-x)/2^n}-1}$$



$$=dfrac{e^{2pi i(z-x)}-1}{e^{2pi i(z-x)/2^n}-1}$$



If $z-x$ is integer $$e^{2pi i(z-x)}=(e^{2pi i})^{z-x}=?$$ provided $e^{2pi i(z-x)/2^n}ne1iff2^nnmid(z-x)$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    We have two cases:




    Case $1$: $z neq x$ (mod. $2^n$) $$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n} = sum_{y=0}^{2^n-1} a^y = frac{1 - a^{2^n}}{1-a}$$
    where $a = Big( e^{2pi i(z-x)/2^n} Big)$ Replacing as @lab bhattacharjee did, we get $0$.




    and




    Case $2$: $z = x$ (mod. $2^n$) $$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n} = sum_{y=0}^{2^n-1} e^0 = 2^n$$







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      What if $2^n$ divide $z-x$
      $endgroup$
      – lab bhattacharjee
      Jan 16 at 15:15










    • $begingroup$
      that's a remarkable note .. should be added as follows $z neq x $ (mod. $2^n$).
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 15:26
















    2












    $begingroup$

    We have two cases:




    Case $1$: $z neq x$ (mod. $2^n$) $$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n} = sum_{y=0}^{2^n-1} a^y = frac{1 - a^{2^n}}{1-a}$$
    where $a = Big( e^{2pi i(z-x)/2^n} Big)$ Replacing as @lab bhattacharjee did, we get $0$.




    and




    Case $2$: $z = x$ (mod. $2^n$) $$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n} = sum_{y=0}^{2^n-1} e^0 = 2^n$$







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      What if $2^n$ divide $z-x$
      $endgroup$
      – lab bhattacharjee
      Jan 16 at 15:15










    • $begingroup$
      that's a remarkable note .. should be added as follows $z neq x $ (mod. $2^n$).
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 15:26














    2












    2








    2





    $begingroup$

    We have two cases:




    Case $1$: $z neq x$ (mod. $2^n$) $$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n} = sum_{y=0}^{2^n-1} a^y = frac{1 - a^{2^n}}{1-a}$$
    where $a = Big( e^{2pi i(z-x)/2^n} Big)$ Replacing as @lab bhattacharjee did, we get $0$.




    and




    Case $2$: $z = x$ (mod. $2^n$) $$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n} = sum_{y=0}^{2^n-1} e^0 = 2^n$$







    share|cite|improve this answer











    $endgroup$



    We have two cases:




    Case $1$: $z neq x$ (mod. $2^n$) $$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n} = sum_{y=0}^{2^n-1} a^y = frac{1 - a^{2^n}}{1-a}$$
    where $a = Big( e^{2pi i(z-x)/2^n} Big)$ Replacing as @lab bhattacharjee did, we get $0$.




    and




    Case $2$: $z = x$ (mod. $2^n$) $$sum_{y=0}^{2^n-1}e^{2pi iy(z-x)/2^n} = sum_{y=0}^{2^n-1} e^0 = 2^n$$








    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 16 at 15:33

























    answered Jan 16 at 14:46









    Ahmad BazziAhmad Bazzi

    8,3282824




    8,3282824












    • $begingroup$
      What if $2^n$ divide $z-x$
      $endgroup$
      – lab bhattacharjee
      Jan 16 at 15:15










    • $begingroup$
      that's a remarkable note .. should be added as follows $z neq x $ (mod. $2^n$).
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 15:26


















    • $begingroup$
      What if $2^n$ divide $z-x$
      $endgroup$
      – lab bhattacharjee
      Jan 16 at 15:15










    • $begingroup$
      that's a remarkable note .. should be added as follows $z neq x $ (mod. $2^n$).
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 15:26
















    $begingroup$
    What if $2^n$ divide $z-x$
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 15:15




    $begingroup$
    What if $2^n$ divide $z-x$
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 15:15












    $begingroup$
    that's a remarkable note .. should be added as follows $z neq x $ (mod. $2^n$).
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 15:26




    $begingroup$
    that's a remarkable note .. should be added as follows $z neq x $ (mod. $2^n$).
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 15:26











    1












    $begingroup$

    $$sum_{y=0}^{2^n-1}(e^{2pi i(z-x)/2^n})^y$$



    $$=dfrac{(e^{2pi i(z-x)/2^n})^{2^n}-1}{e^{2pi i(z-x)/2^n}-1}$$



    $$=dfrac{e^{2pi i(z-x)}-1}{e^{2pi i(z-x)/2^n}-1}$$



    If $z-x$ is integer $$e^{2pi i(z-x)}=(e^{2pi i})^{z-x}=?$$ provided $e^{2pi i(z-x)/2^n}ne1iff2^nnmid(z-x)$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $$sum_{y=0}^{2^n-1}(e^{2pi i(z-x)/2^n})^y$$



      $$=dfrac{(e^{2pi i(z-x)/2^n})^{2^n}-1}{e^{2pi i(z-x)/2^n}-1}$$



      $$=dfrac{e^{2pi i(z-x)}-1}{e^{2pi i(z-x)/2^n}-1}$$



      If $z-x$ is integer $$e^{2pi i(z-x)}=(e^{2pi i})^{z-x}=?$$ provided $e^{2pi i(z-x)/2^n}ne1iff2^nnmid(z-x)$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $$sum_{y=0}^{2^n-1}(e^{2pi i(z-x)/2^n})^y$$



        $$=dfrac{(e^{2pi i(z-x)/2^n})^{2^n}-1}{e^{2pi i(z-x)/2^n}-1}$$



        $$=dfrac{e^{2pi i(z-x)}-1}{e^{2pi i(z-x)/2^n}-1}$$



        If $z-x$ is integer $$e^{2pi i(z-x)}=(e^{2pi i})^{z-x}=?$$ provided $e^{2pi i(z-x)/2^n}ne1iff2^nnmid(z-x)$






        share|cite|improve this answer









        $endgroup$



        $$sum_{y=0}^{2^n-1}(e^{2pi i(z-x)/2^n})^y$$



        $$=dfrac{(e^{2pi i(z-x)/2^n})^{2^n}-1}{e^{2pi i(z-x)/2^n}-1}$$



        $$=dfrac{e^{2pi i(z-x)}-1}{e^{2pi i(z-x)/2^n}-1}$$



        If $z-x$ is integer $$e^{2pi i(z-x)}=(e^{2pi i})^{z-x}=?$$ provided $e^{2pi i(z-x)/2^n}ne1iff2^nnmid(z-x)$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 14:40









        lab bhattacharjeelab bhattacharjee

        226k15157275




        226k15157275






























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