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







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










  • $begingroup$
    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
    55 mins ago










  • $begingroup$
    Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
    $endgroup$
    – robjohn♦
    52 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♦
    48 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♦
    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
    55 mins ago










  • $begingroup$
    Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
    $endgroup$
    – robjohn♦
    52 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♦
    48 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 46 mins ago







cytinus













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.









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♦
    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
    55 mins ago










  • $begingroup$
    Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
    $endgroup$
    – robjohn♦
    52 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♦
    48 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♦
    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
    55 mins ago










  • $begingroup$
    Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
    $endgroup$
    – robjohn♦
    52 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♦
    48 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♦
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












$begingroup$
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
55 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
55 mins ago












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




$begingroup$
Hint: $prodlimits_{m=2}^{n-1}e^{a_m}=e^{sumlimits_{m=2}^{n-1}a_m}$
$endgroup$
– robjohn♦
52 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♦
48 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♦
48 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
    48 mins ago










  • $begingroup$
    @greelious because I messed up... Thanks! :)
    $endgroup$
    – clathratus
    47 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
    40 mins ago












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










  • $begingroup$
    Looks good now.
    $endgroup$
    – greelious
    32 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











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });






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










    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076538%2fclosed-form-of-the-real-portion-of-the-product-of-exponentials%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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
      48 mins ago










    • $begingroup$
      @greelious because I messed up... Thanks! :)
      $endgroup$
      – clathratus
      47 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
      40 mins ago












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










    • $begingroup$
      Looks good now.
      $endgroup$
      – greelious
      32 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
      48 mins ago










    • $begingroup$
      @greelious because I messed up... Thanks! :)
      $endgroup$
      – clathratus
      47 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
      40 mins ago












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










    • $begingroup$
      Looks good now.
      $endgroup$
      – greelious
      32 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 37 mins ago

























    answered 56 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
      48 mins ago










    • $begingroup$
      @greelious because I messed up... Thanks! :)
      $endgroup$
      – clathratus
      47 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
      40 mins ago












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










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














    • 1




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










    • $begingroup$
      @greelious because I messed up... Thanks! :)
      $endgroup$
      – clathratus
      47 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
      40 mins ago












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










    • $begingroup$
      Looks good now.
      $endgroup$
      – greelious
      32 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
    48 mins ago




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












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




    $begingroup$
    @greelious because I messed up... Thanks! :)
    $endgroup$
    – clathratus
    47 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
    40 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
    40 mins ago














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




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












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




    $begingroup$
    Looks good now.
    $endgroup$
    – greelious
    32 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 50 mins ago

























        answered 1 hour ago









        robjohn♦robjohn

        265k27304626




        265k27304626






















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










            draft saved

            draft discarded


















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













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












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
















            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076538%2fclosed-form-of-the-real-portion-of-the-product-of-exponentials%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Polycentropodidae

            Image
            Polycentropodidae Plectrocnemia conspersa Siyentipiko nga pagklasipika Ginhadi-an: Animalia Phylum: Arthropoda Ubosphylum: Hexapoda Klase: Insecta Orden: Trichoptera Labawbanay: Hydropsychoidea Banay: Polycentropodidae Binomial nga ngaran Polycentropodidae An Polycentropodidae [1] in uska familia han Insecta. An Polycentropodidae in nahilalakip ha familia nga Hydropsychoidea, ordo nga Trichoptera, classis nga Insecta, punoan nga Arthropoda, ngan regnum nga Animalia. [1] An familia nga Polycentropodidae in naglalakip hin 660 ka mga species, sumala ha Catalogue of Life [1] . An kladograma hini sumala ha Catalogue of Life [1] : Hydropsychoidea   Polycentropodidae  Adectophylax Antillopsyche Cernotina Cyrnellus Cyrnodes Cyrnopsis Cyrnus Eodipseudopsis Eoneureclipsis Holocentropus Kambaitipsyche Neucentropus
            Read more

            Magento 2 Error message: Invalid state change requested

            Image
            2 1 I've created custom Rest API for adding and removing products from cart. I am able to add and delete products from my cart, but when I log in as a customer, I can't add or delete products from cart. API throws message: Invalid state change requested. How can I solve this? magento2 share | improve this question edited Mar 15 '18 at 13:02 Bare Feet 1,561 1 7 32 asked Mar 15 '18 at 12:50 Prabhu M. Prabhu M. 18 5
            Read more

            Paulmy

            Image
            An Paulmy amo in usa ka komyun ha departamento han Indre-et-Loire ngan ha Rehiyon Sentro ha nasod han Fransya. k h l Mga komyun ha departamento han Indre-et-Loire Abilly  · Ambillou  · Amboise  · Anché  · Antogny-le-Tillac  · Artannes-sur-Indre  · Assay  · Athée-sur-Cher  · Autrèche  · Auzouer-en-Touraine  · Avoine  · Avon-les-Roches  · Avrillé-les-Ponceaux  · Azay-le-Rideau  · Azay-sur-Cher  · Azay-sur-Indre  · Ballan-Miré  · Barrou  · Beaulieu-lès-Loches  · Beaumont-en-Véron  · Beaumont-la-Ronce  · Beaumont-Village  · Benais  · Berthenay  · Betz-le-Château  · Bléré  · Bossay-sur-Claise  · Bossée  · Le Boulay  · Bourgueil  · Bournan  · Boussay  · Braslou  · Braye-sous-Faye  · Braye-sur-Maulne  · Brèches  · Bréhémont  · Bridoré  · Brizay  · Bueil-en-Touraine  · Candes-Saint-Martin  · Cangey  · La Celle-Guenand  · La Celle-Saint-Avant  · Céré-la-Ronde  · Cerelles  · Chambon  ·
            Read more