Closed Form of the Real Portion of the Product of Exponentials












3












$begingroup$


I am wondering if it is possible to express an equation in closed form. I currently have:



$f(n) = prod_{m=2}^{n-1} e^frac{pi i n}{m}$



Where $i$ is the $sqrt{-1}$, which I know it commonly represents but due to the finite product I figured I would add this note for clarity.



Context: I am trying to work towards a solution for Simplify Product of sines, yet so far I have only arrived here.



Some options I have been considering to try and find a closed form include cases such as I only actually care when $n$ is an odd number so discarding even cases is fine if that simplifies it. I am also fine with having the series start a $m=1$ instead of $m=2$ if that somehow simplifies the answer. The other potential saving grace is that I may only care about the real portion of the answer and not the imaginary portion if that simplifies things.



Sorry I am a computer scientist not a mathematician so please let me know if I should adjust the title or the wording of the question for clarity. Any leads or help of any form would be appreciated as I am currently lost.






exponential-function closed-form products







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cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    $endgroup$
    – robjohn♦
    52 mins ago










  • $begingroup$
    Hi, I guess the context is I was trying to work a little more on this, math.stackexchange.com/questions/1689831/…. I'm not in school and this is not homework, I am just a computer scientist who likes to dabble in some of my free time. I was trying to work towards a closed form of the equation linked and so I was breaking it into parts and this is where I got stuck. Its something I look at every few months so even a nudge in any direction would be helpful.
    $endgroup$
    – cytinus
    45 mins ago










  • $begingroup$
    Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
    $endgroup$
    – robjohn♦
    42 mins ago






  • 1




    $begingroup$
    It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
    $endgroup$
    – robjohn♦
    39 mins ago


















3












$begingroup$


I am wondering if it is possible to express an equation in closed form. I currently have:



$f(n) = prod_{m=2}^{n-1} e^frac{pi i n}{m}$



Where $i$ is the $sqrt{-1}$, which I know it commonly represents but due to the finite product I figured I would add this note for clarity.



Context: I am trying to work towards a solution for Simplify Product of sines, yet so far I have only arrived here.



Some options I have been considering to try and find a closed form include cases such as I only actually care when $n$ is an odd number so discarding even cases is fine if that simplifies it. I am also fine with having the series start a $m=1$ instead of $m=2$ if that somehow simplifies the answer. The other potential saving grace is that I may only care about the real portion of the answer and not the imaginary portion if that simplifies things.



Sorry I am a computer scientist not a mathematician so please let me know if I should adjust the title or the wording of the question for clarity. Any leads or help of any form would be appreciated as I am currently lost.






exponential-function closed-form products







share|cite|improve this question









New contributor




cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    $endgroup$
    – robjohn♦
    52 mins ago










  • $begingroup$
    Hi, I guess the context is I was trying to work a little more on this, math.stackexchange.com/questions/1689831/…. I'm not in school and this is not homework, I am just a computer scientist who likes to dabble in some of my free time. I was trying to work towards a closed form of the equation linked and so I was breaking it into parts and this is where I got stuck. Its something I look at every few months so even a nudge in any direction would be helpful.
    $endgroup$
    – cytinus
    45 mins ago










  • $begingroup$
    Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
    $endgroup$
    – robjohn♦
    42 mins ago






  • 1




    $begingroup$
    It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
    $endgroup$
    – robjohn♦
    39 mins ago
















3












3








3





$begingroup$


I am wondering if it is possible to express an equation in closed form. I currently have:



$f(n) = prod_{m=2}^{n-1} e^frac{pi i n}{m}$



Where $i$ is the $sqrt{-1}$, which I know it commonly represents but due to the finite product I figured I would add this note for clarity.



Context: I am trying to work towards a solution for Simplify Product of sines, yet so far I have only arrived here.



Some options I have been considering to try and find a closed form include cases such as I only actually care when $n$ is an odd number so discarding even cases is fine if that simplifies it. I am also fine with having the series start a $m=1$ instead of $m=2$ if that somehow simplifies the answer. The other potential saving grace is that I may only care about the real portion of the answer and not the imaginary portion if that simplifies things.



Sorry I am a computer scientist not a mathematician so please let me know if I should adjust the title or the wording of the question for clarity. Any leads or help of any form would be appreciated as I am currently lost.






exponential-function closed-form products







share|cite|improve this question









New contributor




cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I am wondering if it is possible to express an equation in closed form. I currently have:



$f(n) = prod_{m=2}^{n-1} e^frac{pi i n}{m}$



Where $i$ is the $sqrt{-1}$, which I know it commonly represents but due to the finite product I figured I would add this note for clarity.



Context: I am trying to work towards a solution for Simplify Product of sines, yet so far I have only arrived here.



Some options I have been considering to try and find a closed form include cases such as I only actually care when $n$ is an odd number so discarding even cases is fine if that simplifies it. I am also fine with having the series start a $m=1$ instead of $m=2$ if that somehow simplifies the answer. The other potential saving grace is that I may only care about the real portion of the answer and not the imaginary portion if that simplifies things.



Sorry I am a computer scientist not a mathematician so please let me know if I should adjust the title or the wording of the question for clarity. Any leads or help of any form would be appreciated as I am currently lost.






exponential-function closed-form products




exponential-function closed-form products






share|cite|improve this question









New contributor




cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 36 mins ago







cytinus













New contributor




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Check out our Code of Conduct.









asked 1 hour ago









cytinuscytinus

1164




1164




New contributor




cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    $endgroup$
    – robjohn♦
    52 mins ago










  • $begingroup$
    Hi, I guess the context is I was trying to work a little more on this, math.stackexchange.com/questions/1689831/…. I'm not in school and this is not homework, I am just a computer scientist who likes to dabble in some of my free time. I was trying to work towards a closed form of the equation linked and so I was breaking it into parts and this is where I got stuck. Its something I look at every few months so even a nudge in any direction would be helpful.
    $endgroup$
    – cytinus
    45 mins ago










  • $begingroup$
    Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
    $endgroup$
    – robjohn♦
    42 mins ago






  • 1




    $begingroup$
    It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
    $endgroup$
    – robjohn♦
    39 mins ago




















  • $begingroup$
    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    $endgroup$
    – robjohn♦
    52 mins ago










  • $begingroup$
    Hi, I guess the context is I was trying to work a little more on this, math.stackexchange.com/questions/1689831/…. I'm not in school and this is not homework, I am just a computer scientist who likes to dabble in some of my free time. I was trying to work towards a closed form of the equation linked and so I was breaking it into parts and this is where I got stuck. Its something I look at every few months so even a nudge in any direction would be helpful.
    $endgroup$
    – cytinus
    45 mins ago










  • $begingroup$
    Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
    $endgroup$
    – robjohn♦
    42 mins ago






  • 1




    $begingroup$
    It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
    $endgroup$
    – robjohn♦
    39 mins ago


















$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
52 mins ago




$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
52 mins ago












$begingroup$
Hi, I guess the context is I was trying to work a little more on this, math.stackexchange.com/questions/1689831/…. I'm not in school and this is not homework, I am just a computer scientist who likes to dabble in some of my free time. I was trying to work towards a closed form of the equation linked and so I was breaking it into parts and this is where I got stuck. Its something I look at every few months so even a nudge in any direction would be helpful.
$endgroup$
– cytinus
45 mins ago




$begingroup$
Hi, I guess the context is I was trying to work a little more on this, math.stackexchange.com/questions/1689831/…. I'm not in school and this is not homework, I am just a computer scientist who likes to dabble in some of my free time. I was trying to work towards a closed form of the equation linked and so I was breaking it into parts and this is where I got stuck. Its something I look at every few months so even a nudge in any direction would be helpful.
$endgroup$
– cytinus
45 mins ago












$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
42 mins ago




$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
42 mins ago




1




1




$begingroup$
It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
$endgroup$
– robjohn♦
39 mins ago






$begingroup$
It would be helpful to add that context to the question itself, rather than in a comment. I have undeleted my answer.
$endgroup$
– robjohn♦
39 mins ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

$$begin{align}
f(n)=&prod_{m=2}^{n-1}expfrac{ipi n}{m}\
&=expleft[sum_{m=2}^{n-1}frac{ipi n}{m}right]\
&=expleft[ipi nsum_{m=2}^{n-1}frac{1}{m}right]\
end{align}$$
Recalling the definition of the harmonic numbers:
$$H_n=sum_{m=1}^nfrac1m$$
We have that
$$f(n)=expleft[ipi n(H_{n-1}-1)right]$$
Then using $e^{itheta}=costheta+isintheta$,
$$f(n)=cosleft[pi n(H_{n-1}-1)right]+isinleft[pi n(H_{n-1}-1)right]$$
So
$$text{Re}f(n)=cosleft[pi n(H_{n-1}-1)right]$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
    $endgroup$
    – greelious
    38 mins ago










  • $begingroup$
    @greelious because I messed up... Thanks! :)
    $endgroup$
    – clathratus
    37 mins ago










  • $begingroup$
    I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
    $endgroup$
    – greelious
    30 mins ago












  • $begingroup$
    @greelious okay look now
    $endgroup$
    – clathratus
    26 mins ago










  • $begingroup$
    Looks good now.
    $endgroup$
    – greelious
    23 mins ago



















2












$begingroup$

$$
begin{align}
prod_{m=2}^{n-1}e^{frac{pi in}m}
&=e^{pi in(H_{n-1}-1)}\
&=(-1)^ne^{pi inH_{n-1}}\[9pt]
&=(-1)^{n-1}e^{pi inH_n}tag1
end{align}
$$
where $H_n$ is the $n^text{th}$ Harmonic Number.
$$
H_nsimlog(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}tag2
$$
and $gamma$ is the Euler-Mascheroni constant.



The real portion of $(1)$ is
$$
operatorname{Re}left(prod_{m=2}^{n-1}e^{frac{pi in}m}right)=(-1)^{n-1}cosleft(pi nH_nright)tag3
$$






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









    3












    $begingroup$

    $$begin{align}
    f(n)=&prod_{m=2}^{n-1}expfrac{ipi n}{m}\
    &=expleft[sum_{m=2}^{n-1}frac{ipi n}{m}right]\
    &=expleft[ipi nsum_{m=2}^{n-1}frac{1}{m}right]\
    end{align}$$
    Recalling the definition of the harmonic numbers:
    $$H_n=sum_{m=1}^nfrac1m$$
    We have that
    $$f(n)=expleft[ipi n(H_{n-1}-1)right]$$
    Then using $e^{itheta}=costheta+isintheta$,
    $$f(n)=cosleft[pi n(H_{n-1}-1)right]+isinleft[pi n(H_{n-1}-1)right]$$
    So
    $$text{Re}f(n)=cosleft[pi n(H_{n-1}-1)right]$$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
      $endgroup$
      – greelious
      38 mins ago










    • $begingroup$
      @greelious because I messed up... Thanks! :)
      $endgroup$
      – clathratus
      37 mins ago










    • $begingroup$
      I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
      $endgroup$
      – greelious
      30 mins ago












    • $begingroup$
      @greelious okay look now
      $endgroup$
      – clathratus
      26 mins ago










    • $begingroup$
      Looks good now.
      $endgroup$
      – greelious
      23 mins ago
















    3












    $begingroup$

    $$begin{align}
    f(n)=&prod_{m=2}^{n-1}expfrac{ipi n}{m}\
    &=expleft[sum_{m=2}^{n-1}frac{ipi n}{m}right]\
    &=expleft[ipi nsum_{m=2}^{n-1}frac{1}{m}right]\
    end{align}$$
    Recalling the definition of the harmonic numbers:
    $$H_n=sum_{m=1}^nfrac1m$$
    We have that
    $$f(n)=expleft[ipi n(H_{n-1}-1)right]$$
    Then using $e^{itheta}=costheta+isintheta$,
    $$f(n)=cosleft[pi n(H_{n-1}-1)right]+isinleft[pi n(H_{n-1}-1)right]$$
    So
    $$text{Re}f(n)=cosleft[pi n(H_{n-1}-1)right]$$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
      $endgroup$
      – greelious
      38 mins ago










    • $begingroup$
      @greelious because I messed up... Thanks! :)
      $endgroup$
      – clathratus
      37 mins ago










    • $begingroup$
      I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
      $endgroup$
      – greelious
      30 mins ago












    • $begingroup$
      @greelious okay look now
      $endgroup$
      – clathratus
      26 mins ago










    • $begingroup$
      Looks good now.
      $endgroup$
      – greelious
      23 mins ago














    3












    3








    3





    $begingroup$

    $$begin{align}
    f(n)=&prod_{m=2}^{n-1}expfrac{ipi n}{m}\
    &=expleft[sum_{m=2}^{n-1}frac{ipi n}{m}right]\
    &=expleft[ipi nsum_{m=2}^{n-1}frac{1}{m}right]\
    end{align}$$
    Recalling the definition of the harmonic numbers:
    $$H_n=sum_{m=1}^nfrac1m$$
    We have that
    $$f(n)=expleft[ipi n(H_{n-1}-1)right]$$
    Then using $e^{itheta}=costheta+isintheta$,
    $$f(n)=cosleft[pi n(H_{n-1}-1)right]+isinleft[pi n(H_{n-1}-1)right]$$
    So
    $$text{Re}f(n)=cosleft[pi n(H_{n-1}-1)right]$$






    share|cite|improve this answer











    $endgroup$



    $$begin{align}
    f(n)=&prod_{m=2}^{n-1}expfrac{ipi n}{m}\
    &=expleft[sum_{m=2}^{n-1}frac{ipi n}{m}right]\
    &=expleft[ipi nsum_{m=2}^{n-1}frac{1}{m}right]\
    end{align}$$
    Recalling the definition of the harmonic numbers:
    $$H_n=sum_{m=1}^nfrac1m$$
    We have that
    $$f(n)=expleft[ipi n(H_{n-1}-1)right]$$
    Then using $e^{itheta}=costheta+isintheta$,
    $$f(n)=cosleft[pi n(H_{n-1}-1)right]+isinleft[pi n(H_{n-1}-1)right]$$
    So
    $$text{Re}f(n)=cosleft[pi n(H_{n-1}-1)right]$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 27 mins ago

























    answered 46 mins ago









    clathratusclathratus

    3,606332




    3,606332








    • 1




      $begingroup$
      Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
      $endgroup$
      – greelious
      38 mins ago










    • $begingroup$
      @greelious because I messed up... Thanks! :)
      $endgroup$
      – clathratus
      37 mins ago










    • $begingroup$
      I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
      $endgroup$
      – greelious
      30 mins ago












    • $begingroup$
      @greelious okay look now
      $endgroup$
      – clathratus
      26 mins ago










    • $begingroup$
      Looks good now.
      $endgroup$
      – greelious
      23 mins ago














    • 1




      $begingroup$
      Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
      $endgroup$
      – greelious
      38 mins ago










    • $begingroup$
      @greelious because I messed up... Thanks! :)
      $endgroup$
      – clathratus
      37 mins ago










    • $begingroup$
      I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
      $endgroup$
      – greelious
      30 mins ago












    • $begingroup$
      @greelious okay look now
      $endgroup$
      – clathratus
      26 mins ago










    • $begingroup$
      Looks good now.
      $endgroup$
      – greelious
      23 mins ago








    1




    1




    $begingroup$
    Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
    $endgroup$
    – greelious
    38 mins ago




    $begingroup$
    Why is $expleft[ipi n(H_{n-1}-1)right]=exp(ipi n)exp(H_{n-1}-1)$?
    $endgroup$
    – greelious
    38 mins ago












    $begingroup$
    @greelious because I messed up... Thanks! :)
    $endgroup$
    – clathratus
    37 mins ago




    $begingroup$
    @greelious because I messed up... Thanks! :)
    $endgroup$
    – clathratus
    37 mins ago












    $begingroup$
    I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
    $endgroup$
    – greelious
    30 mins ago






    $begingroup$
    I think it is still not correct: $expleft[ipi n(H_{n-1}-1)right]=exp(i pi n H_{n-1})exp(-i pi n)=exp(i pi n H_{n-1})(-1)^n$
    $endgroup$
    – greelious
    30 mins ago














    $begingroup$
    @greelious okay look now
    $endgroup$
    – clathratus
    26 mins ago




    $begingroup$
    @greelious okay look now
    $endgroup$
    – clathratus
    26 mins ago












    $begingroup$
    Looks good now.
    $endgroup$
    – greelious
    23 mins ago




    $begingroup$
    Looks good now.
    $endgroup$
    – greelious
    23 mins ago











    2












    $begingroup$

    $$
    begin{align}
    prod_{m=2}^{n-1}e^{frac{pi in}m}
    &=e^{pi in(H_{n-1}-1)}\
    &=(-1)^ne^{pi inH_{n-1}}\[9pt]
    &=(-1)^{n-1}e^{pi inH_n}tag1
    end{align}
    $$
    where $H_n$ is the $n^text{th}$ Harmonic Number.
    $$
    H_nsimlog(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}tag2
    $$
    and $gamma$ is the Euler-Mascheroni constant.



    The real portion of $(1)$ is
    $$
    operatorname{Re}left(prod_{m=2}^{n-1}e^{frac{pi in}m}right)=(-1)^{n-1}cosleft(pi nH_nright)tag3
    $$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      $$
      begin{align}
      prod_{m=2}^{n-1}e^{frac{pi in}m}
      &=e^{pi in(H_{n-1}-1)}\
      &=(-1)^ne^{pi inH_{n-1}}\[9pt]
      &=(-1)^{n-1}e^{pi inH_n}tag1
      end{align}
      $$
      where $H_n$ is the $n^text{th}$ Harmonic Number.
      $$
      H_nsimlog(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}tag2
      $$
      and $gamma$ is the Euler-Mascheroni constant.



      The real portion of $(1)$ is
      $$
      operatorname{Re}left(prod_{m=2}^{n-1}e^{frac{pi in}m}right)=(-1)^{n-1}cosleft(pi nH_nright)tag3
      $$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        $$
        begin{align}
        prod_{m=2}^{n-1}e^{frac{pi in}m}
        &=e^{pi in(H_{n-1}-1)}\
        &=(-1)^ne^{pi inH_{n-1}}\[9pt]
        &=(-1)^{n-1}e^{pi inH_n}tag1
        end{align}
        $$
        where $H_n$ is the $n^text{th}$ Harmonic Number.
        $$
        H_nsimlog(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}tag2
        $$
        and $gamma$ is the Euler-Mascheroni constant.



        The real portion of $(1)$ is
        $$
        operatorname{Re}left(prod_{m=2}^{n-1}e^{frac{pi in}m}right)=(-1)^{n-1}cosleft(pi nH_nright)tag3
        $$






        share|cite|improve this answer











        $endgroup$



        $$
        begin{align}
        prod_{m=2}^{n-1}e^{frac{pi in}m}
        &=e^{pi in(H_{n-1}-1)}\
        &=(-1)^ne^{pi inH_{n-1}}\[9pt]
        &=(-1)^{n-1}e^{pi inH_n}tag1
        end{align}
        $$
        where $H_n$ is the $n^text{th}$ Harmonic Number.
        $$
        H_nsimlog(n)+gamma+frac1{2n}-frac1{12n^2}+frac1{120n^4}-frac1{252n^6}+frac1{240n^8}-frac1{132n^{10}}tag2
        $$
        and $gamma$ is the Euler-Mascheroni constant.



        The real portion of $(1)$ is
        $$
        operatorname{Re}left(prod_{m=2}^{n-1}e^{frac{pi in}m}right)=(-1)^{n-1}cosleft(pi nH_nright)tag3
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 40 mins ago

























        answered 1 hour ago









        robjohn♦robjohn

        265k27304626




        265k27304626






















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