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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  • $begingroup$
    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♦
    1 hour ago






  • 1




    $begingroup$
    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
    1 hour ago










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






  • 1




    $begingroup$
    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
    1 hour ago










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







cytinus













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asked 2 hours ago









cytinuscytinus

1164




1164




New contributor




cytinus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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♦
    1 hour ago






  • 1




    $begingroup$
    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
    1 hour ago










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






  • 1




    $begingroup$
    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
    1 hour ago










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




1




1




$begingroup$
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
1 hour 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
1 hour ago












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




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












2 Answers
2






active

oldest

votes


















3












$begingroup$

$$
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











$endgroup$





















    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
      1 hour ago










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












    • $begingroup$
      @greelious okay look now
      $endgroup$
      – clathratus
      1 hour ago










    • $begingroup$
      Looks good now.
      $endgroup$
      – greelious
      1 hour ago











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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}
    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











    $endgroup$


















      3












      $begingroup$

      $$
      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











      $endgroup$
















        3












        3








        3





        $begingroup$

        $$
        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 1 hour ago

























        answered 2 hours ago









        robjohn♦robjohn

        265k27304626




        265k27304626























            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
              1 hour ago










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












            • $begingroup$
              @greelious okay look now
              $endgroup$
              – clathratus
              1 hour ago










            • $begingroup$
              Looks good now.
              $endgroup$
              – greelious
              1 hour 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
              1 hour ago










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












            • $begingroup$
              @greelious okay look now
              $endgroup$
              – clathratus
              1 hour ago










            • $begingroup$
              Looks good now.
              $endgroup$
              – greelious
              1 hour 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 1 hour ago

























            answered 1 hour ago









            clathratusclathratus

            3,616332




            3,616332








            • 1




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










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












            • $begingroup$
              @greelious okay look now
              $endgroup$
              – clathratus
              1 hour ago










            • $begingroup$
              Looks good now.
              $endgroup$
              – greelious
              1 hour ago














            • 1




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










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












            • $begingroup$
              @greelious okay look now
              $endgroup$
              – clathratus
              1 hour ago










            • $begingroup$
              Looks good now.
              $endgroup$
              – greelious
              1 hour ago








            1




            1




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




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












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




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














            $begingroup$
            @greelious okay look now
            $endgroup$
            – clathratus
            1 hour ago




            $begingroup$
            @greelious okay look now
            $endgroup$
            – clathratus
            1 hour ago












            $begingroup$
            Looks good now.
            $endgroup$
            – greelious
            1 hour ago




            $begingroup$
            Looks good now.
            $endgroup$
            – greelious
            1 hour ago










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            cytinus is a new contributor. Be nice, and check out our Code of Conduct.
















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