Is there a name for this series?












3












$begingroup$


I am doing a presentation on nonlinear optics and I ran into a paper that uses a series to describe a wave equation of a system. The paper will be linked below and the series is:



$sum_{n=1}^{infty}frac{(-i)^{n-1}}{n!}k^n$



I mostly am looking for what i should call this other than "series." The equations are presented on pg. 1-2, eqns. 1a, 1b, and 1c (all are the same thing for 3 different waves).



http://www.few.vu.nl/~switte/papers/OPCPA_review_published_early_edition.pdf



EDIT: The full equation is



$frac{partial A}{partial z}+sum_{n=1}^{infty}frac{(-i)^{n-1}}{n!}k^n frac{partial^{n} A}{partial t^{n}}=-i frac{chi^{(2)}omega}{2nc}AA^{*}e^{-iDelta textbf{k}cdot textbf{z}}$



so it may just be simpler to discuss it in terms of what each portion does rather than by mathematical terminology..






sequences-and-series power-series







share|cite|improve this question









New contributor




celani99 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    After checking the paper, the formula is much more complicated than that. Voting to close.
    $endgroup$
    – Yves Daoust
    15 hours ago












  • $begingroup$
    @YvesDaoust Should I delete the question instead?
    $endgroup$
    – celani99
    14 hours ago










  • $begingroup$
    You still have the option of updating the question to ask for clarification about the formula.
    $endgroup$
    – Yves Daoust
    14 hours ago










  • $begingroup$
    @YvesDaoust Is that edit sufficient, or should I add any other information?
    $endgroup$
    – celani99
    14 hours ago










  • $begingroup$
    $^{(n)}$, not $^n$ !
    $endgroup$
    – Yves Daoust
    13 hours ago
















3












$begingroup$


I am doing a presentation on nonlinear optics and I ran into a paper that uses a series to describe a wave equation of a system. The paper will be linked below and the series is:



$sum_{n=1}^{infty}frac{(-i)^{n-1}}{n!}k^n$



I mostly am looking for what i should call this other than "series." The equations are presented on pg. 1-2, eqns. 1a, 1b, and 1c (all are the same thing for 3 different waves).



http://www.few.vu.nl/~switte/papers/OPCPA_review_published_early_edition.pdf



EDIT: The full equation is



$frac{partial A}{partial z}+sum_{n=1}^{infty}frac{(-i)^{n-1}}{n!}k^n frac{partial^{n} A}{partial t^{n}}=-i frac{chi^{(2)}omega}{2nc}AA^{*}e^{-iDelta textbf{k}cdot textbf{z}}$



so it may just be simpler to discuss it in terms of what each portion does rather than by mathematical terminology..






sequences-and-series power-series







share|cite|improve this question









New contributor




celani99 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    After checking the paper, the formula is much more complicated than that. Voting to close.
    $endgroup$
    – Yves Daoust
    15 hours ago












  • $begingroup$
    @YvesDaoust Should I delete the question instead?
    $endgroup$
    – celani99
    14 hours ago










  • $begingroup$
    You still have the option of updating the question to ask for clarification about the formula.
    $endgroup$
    – Yves Daoust
    14 hours ago










  • $begingroup$
    @YvesDaoust Is that edit sufficient, or should I add any other information?
    $endgroup$
    – celani99
    14 hours ago










  • $begingroup$
    $^{(n)}$, not $^n$ !
    $endgroup$
    – Yves Daoust
    13 hours ago














3












3








3





$begingroup$


I am doing a presentation on nonlinear optics and I ran into a paper that uses a series to describe a wave equation of a system. The paper will be linked below and the series is:



$sum_{n=1}^{infty}frac{(-i)^{n-1}}{n!}k^n$



I mostly am looking for what i should call this other than "series." The equations are presented on pg. 1-2, eqns. 1a, 1b, and 1c (all are the same thing for 3 different waves).



http://www.few.vu.nl/~switte/papers/OPCPA_review_published_early_edition.pdf



EDIT: The full equation is



$frac{partial A}{partial z}+sum_{n=1}^{infty}frac{(-i)^{n-1}}{n!}k^n frac{partial^{n} A}{partial t^{n}}=-i frac{chi^{(2)}omega}{2nc}AA^{*}e^{-iDelta textbf{k}cdot textbf{z}}$



so it may just be simpler to discuss it in terms of what each portion does rather than by mathematical terminology..






sequences-and-series power-series







share|cite|improve this question









New contributor




celani99 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 doing a presentation on nonlinear optics and I ran into a paper that uses a series to describe a wave equation of a system. The paper will be linked below and the series is:



$sum_{n=1}^{infty}frac{(-i)^{n-1}}{n!}k^n$



I mostly am looking for what i should call this other than "series." The equations are presented on pg. 1-2, eqns. 1a, 1b, and 1c (all are the same thing for 3 different waves).



http://www.few.vu.nl/~switte/papers/OPCPA_review_published_early_edition.pdf



EDIT: The full equation is



$frac{partial A}{partial z}+sum_{n=1}^{infty}frac{(-i)^{n-1}}{n!}k^n frac{partial^{n} A}{partial t^{n}}=-i frac{chi^{(2)}omega}{2nc}AA^{*}e^{-iDelta textbf{k}cdot textbf{z}}$



so it may just be simpler to discuss it in terms of what each portion does rather than by mathematical terminology..






sequences-and-series power-series




sequences-and-series power-series






share|cite|improve this question









New contributor




celani99 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




celani99 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 14 hours ago







celani99













New contributor




celani99 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 15 hours ago









celani99celani99

193




193




New contributor




celani99 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





celani99 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






celani99 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    After checking the paper, the formula is much more complicated than that. Voting to close.
    $endgroup$
    – Yves Daoust
    15 hours ago












  • $begingroup$
    @YvesDaoust Should I delete the question instead?
    $endgroup$
    – celani99
    14 hours ago










  • $begingroup$
    You still have the option of updating the question to ask for clarification about the formula.
    $endgroup$
    – Yves Daoust
    14 hours ago










  • $begingroup$
    @YvesDaoust Is that edit sufficient, or should I add any other information?
    $endgroup$
    – celani99
    14 hours ago










  • $begingroup$
    $^{(n)}$, not $^n$ !
    $endgroup$
    – Yves Daoust
    13 hours ago














  • 1




    $begingroup$
    After checking the paper, the formula is much more complicated than that. Voting to close.
    $endgroup$
    – Yves Daoust
    15 hours ago












  • $begingroup$
    @YvesDaoust Should I delete the question instead?
    $endgroup$
    – celani99
    14 hours ago










  • $begingroup$
    You still have the option of updating the question to ask for clarification about the formula.
    $endgroup$
    – Yves Daoust
    14 hours ago










  • $begingroup$
    @YvesDaoust Is that edit sufficient, or should I add any other information?
    $endgroup$
    – celani99
    14 hours ago










  • $begingroup$
    $^{(n)}$, not $^n$ !
    $endgroup$
    – Yves Daoust
    13 hours ago








1




1




$begingroup$
After checking the paper, the formula is much more complicated than that. Voting to close.
$endgroup$
– Yves Daoust
15 hours ago






$begingroup$
After checking the paper, the formula is much more complicated than that. Voting to close.
$endgroup$
– Yves Daoust
15 hours ago














$begingroup$
@YvesDaoust Should I delete the question instead?
$endgroup$
– celani99
14 hours ago




$begingroup$
@YvesDaoust Should I delete the question instead?
$endgroup$
– celani99
14 hours ago












$begingroup$
You still have the option of updating the question to ask for clarification about the formula.
$endgroup$
– Yves Daoust
14 hours ago




$begingroup$
You still have the option of updating the question to ask for clarification about the formula.
$endgroup$
– Yves Daoust
14 hours ago












$begingroup$
@YvesDaoust Is that edit sufficient, or should I add any other information?
$endgroup$
– celani99
14 hours ago




$begingroup$
@YvesDaoust Is that edit sufficient, or should I add any other information?
$endgroup$
– celani99
14 hours ago












$begingroup$
$^{(n)}$, not $^n$ !
$endgroup$
– Yves Daoust
13 hours ago




$begingroup$
$^{(n)}$, not $^n$ !
$endgroup$
– Yves Daoust
13 hours ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

$$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$




This can be rewritten as



$$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    $$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{k}-1).$$ So I would say yes it is known and there is a name for it -- exponential function.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Whyever this got a downvote... If you are not satisfied with the answer, you should comment and tell me what is wrong with it.
      $endgroup$
      – James
      15 hours ago










    • $begingroup$
      @celani99 I saw the OPCPA article, are you sure $k^n=k^{(n)}$? "The k(n) - terms are the nth order dispersion coefficients of the medium" I think k is depends on somehow from n.
      $endgroup$
      – JV.Stalker
      14 hours ago










    • $begingroup$
      @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
      $endgroup$
      – celani99
      14 hours ago








    • 4




      $begingroup$
      Isn't it $e^{-ik}$ rather than $e^k$?
      $endgroup$
      – Acccumulation
      13 hours ago










    • $begingroup$
      @Acccumulation: amazing so many people upvoted without noticing.
      $endgroup$
      – Yves Daoust
      12 hours ago



















    1












    $begingroup$

    Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.



    $sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      The formula in the question is with $k^n$, not $k_n$. And that in the paper is with $k^{(n)}$, which is quite different.
      $endgroup$
      – Yves Daoust
      14 hours ago






    • 1




      $begingroup$
      @YvesDaoust, you are right, I am just would have liked to help to celani99.
      $endgroup$
      – JV.Stalker
      14 hours ago










    • $begingroup$
      No, this is like saying that the name of the function $cos$ is "trigonometric function". Exponential generating functions are just a type of generating function. These are a collection of methods for manipulating series by treating the terms of the series as coefficients in something like a Taylor series and then manipulating the function represented by that series, instead of the original series of numbers.
      $endgroup$
      – David Richerby
      13 hours ago








    • 1




      $begingroup$
      @DavidRicherby I guess you are right, but don't be too hard on JV. Recognizing this as an EGF could be useful in the process of identifying the function, just like noticing that something is a trigonometric function can be useful in precisely determining which function it is.
      $endgroup$
      – Tempestas Ludi
      12 hours ago











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    $$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$




    This can be rewritten as



    $$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.






    share|cite|improve this answer











    $endgroup$


















      5












      $begingroup$

      $$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$




      This can be rewritten as



      $$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.






      share|cite|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        $$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$




        This can be rewritten as



        $$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.






        share|cite|improve this answer











        $endgroup$



        $$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$




        This can be rewritten as



        $$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 14 hours ago

























        answered 15 hours ago









        Yves DaoustYves Daoust

        128k675227




        128k675227























            3












            $begingroup$

            $$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{k}-1).$$ So I would say yes it is known and there is a name for it -- exponential function.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Whyever this got a downvote... If you are not satisfied with the answer, you should comment and tell me what is wrong with it.
              $endgroup$
              – James
              15 hours ago










            • $begingroup$
              @celani99 I saw the OPCPA article, are you sure $k^n=k^{(n)}$? "The k(n) - terms are the nth order dispersion coefficients of the medium" I think k is depends on somehow from n.
              $endgroup$
              – JV.Stalker
              14 hours ago










            • $begingroup$
              @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
              $endgroup$
              – celani99
              14 hours ago








            • 4




              $begingroup$
              Isn't it $e^{-ik}$ rather than $e^k$?
              $endgroup$
              – Acccumulation
              13 hours ago










            • $begingroup$
              @Acccumulation: amazing so many people upvoted without noticing.
              $endgroup$
              – Yves Daoust
              12 hours ago
















            3












            $begingroup$

            $$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{k}-1).$$ So I would say yes it is known and there is a name for it -- exponential function.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Whyever this got a downvote... If you are not satisfied with the answer, you should comment and tell me what is wrong with it.
              $endgroup$
              – James
              15 hours ago










            • $begingroup$
              @celani99 I saw the OPCPA article, are you sure $k^n=k^{(n)}$? "The k(n) - terms are the nth order dispersion coefficients of the medium" I think k is depends on somehow from n.
              $endgroup$
              – JV.Stalker
              14 hours ago










            • $begingroup$
              @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
              $endgroup$
              – celani99
              14 hours ago








            • 4




              $begingroup$
              Isn't it $e^{-ik}$ rather than $e^k$?
              $endgroup$
              – Acccumulation
              13 hours ago










            • $begingroup$
              @Acccumulation: amazing so many people upvoted without noticing.
              $endgroup$
              – Yves Daoust
              12 hours ago














            3












            3








            3





            $begingroup$

            $$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{k}-1).$$ So I would say yes it is known and there is a name for it -- exponential function.






            share|cite|improve this answer











            $endgroup$



            $$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{k}-1).$$ So I would say yes it is known and there is a name for it -- exponential function.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 15 hours ago

























            answered 15 hours ago









            JamesJames

            1,547217




            1,547217








            • 1




              $begingroup$
              Whyever this got a downvote... If you are not satisfied with the answer, you should comment and tell me what is wrong with it.
              $endgroup$
              – James
              15 hours ago










            • $begingroup$
              @celani99 I saw the OPCPA article, are you sure $k^n=k^{(n)}$? "The k(n) - terms are the nth order dispersion coefficients of the medium" I think k is depends on somehow from n.
              $endgroup$
              – JV.Stalker
              14 hours ago










            • $begingroup$
              @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
              $endgroup$
              – celani99
              14 hours ago








            • 4




              $begingroup$
              Isn't it $e^{-ik}$ rather than $e^k$?
              $endgroup$
              – Acccumulation
              13 hours ago










            • $begingroup$
              @Acccumulation: amazing so many people upvoted without noticing.
              $endgroup$
              – Yves Daoust
              12 hours ago














            • 1




              $begingroup$
              Whyever this got a downvote... If you are not satisfied with the answer, you should comment and tell me what is wrong with it.
              $endgroup$
              – James
              15 hours ago










            • $begingroup$
              @celani99 I saw the OPCPA article, are you sure $k^n=k^{(n)}$? "The k(n) - terms are the nth order dispersion coefficients of the medium" I think k is depends on somehow from n.
              $endgroup$
              – JV.Stalker
              14 hours ago










            • $begingroup$
              @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
              $endgroup$
              – celani99
              14 hours ago








            • 4




              $begingroup$
              Isn't it $e^{-ik}$ rather than $e^k$?
              $endgroup$
              – Acccumulation
              13 hours ago










            • $begingroup$
              @Acccumulation: amazing so many people upvoted without noticing.
              $endgroup$
              – Yves Daoust
              12 hours ago








            1




            1




            $begingroup$
            Whyever this got a downvote... If you are not satisfied with the answer, you should comment and tell me what is wrong with it.
            $endgroup$
            – James
            15 hours ago




            $begingroup$
            Whyever this got a downvote... If you are not satisfied with the answer, you should comment and tell me what is wrong with it.
            $endgroup$
            – James
            15 hours ago












            $begingroup$
            @celani99 I saw the OPCPA article, are you sure $k^n=k^{(n)}$? "The k(n) - terms are the nth order dispersion coefficients of the medium" I think k is depends on somehow from n.
            $endgroup$
            – JV.Stalker
            14 hours ago




            $begingroup$
            @celani99 I saw the OPCPA article, are you sure $k^n=k^{(n)}$? "The k(n) - terms are the nth order dispersion coefficients of the medium" I think k is depends on somehow from n.
            $endgroup$
            – JV.Stalker
            14 hours ago












            $begingroup$
            @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
            $endgroup$
            – celani99
            14 hours ago






            $begingroup$
            @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
            $endgroup$
            – celani99
            14 hours ago






            4




            4




            $begingroup$
            Isn't it $e^{-ik}$ rather than $e^k$?
            $endgroup$
            – Acccumulation
            13 hours ago




            $begingroup$
            Isn't it $e^{-ik}$ rather than $e^k$?
            $endgroup$
            – Acccumulation
            13 hours ago












            $begingroup$
            @Acccumulation: amazing so many people upvoted without noticing.
            $endgroup$
            – Yves Daoust
            12 hours ago




            $begingroup$
            @Acccumulation: amazing so many people upvoted without noticing.
            $endgroup$
            – Yves Daoust
            12 hours ago











            1












            $begingroup$

            Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.



            $sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              The formula in the question is with $k^n$, not $k_n$. And that in the paper is with $k^{(n)}$, which is quite different.
              $endgroup$
              – Yves Daoust
              14 hours ago






            • 1




              $begingroup$
              @YvesDaoust, you are right, I am just would have liked to help to celani99.
              $endgroup$
              – JV.Stalker
              14 hours ago










            • $begingroup$
              No, this is like saying that the name of the function $cos$ is "trigonometric function". Exponential generating functions are just a type of generating function. These are a collection of methods for manipulating series by treating the terms of the series as coefficients in something like a Taylor series and then manipulating the function represented by that series, instead of the original series of numbers.
              $endgroup$
              – David Richerby
              13 hours ago








            • 1




              $begingroup$
              @DavidRicherby I guess you are right, but don't be too hard on JV. Recognizing this as an EGF could be useful in the process of identifying the function, just like noticing that something is a trigonometric function can be useful in precisely determining which function it is.
              $endgroup$
              – Tempestas Ludi
              12 hours ago
















            1












            $begingroup$

            Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.



            $sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              The formula in the question is with $k^n$, not $k_n$. And that in the paper is with $k^{(n)}$, which is quite different.
              $endgroup$
              – Yves Daoust
              14 hours ago






            • 1




              $begingroup$
              @YvesDaoust, you are right, I am just would have liked to help to celani99.
              $endgroup$
              – JV.Stalker
              14 hours ago










            • $begingroup$
              No, this is like saying that the name of the function $cos$ is "trigonometric function". Exponential generating functions are just a type of generating function. These are a collection of methods for manipulating series by treating the terms of the series as coefficients in something like a Taylor series and then manipulating the function represented by that series, instead of the original series of numbers.
              $endgroup$
              – David Richerby
              13 hours ago








            • 1




              $begingroup$
              @DavidRicherby I guess you are right, but don't be too hard on JV. Recognizing this as an EGF could be useful in the process of identifying the function, just like noticing that something is a trigonometric function can be useful in precisely determining which function it is.
              $endgroup$
              – Tempestas Ludi
              12 hours ago














            1












            1








            1





            $begingroup$

            Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.



            $sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$






            share|cite|improve this answer











            $endgroup$



            Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.



            $sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 14 hours ago

























            answered 14 hours ago









            JV.StalkerJV.Stalker

            86649




            86649








            • 1




              $begingroup$
              The formula in the question is with $k^n$, not $k_n$. And that in the paper is with $k^{(n)}$, which is quite different.
              $endgroup$
              – Yves Daoust
              14 hours ago






            • 1




              $begingroup$
              @YvesDaoust, you are right, I am just would have liked to help to celani99.
              $endgroup$
              – JV.Stalker
              14 hours ago










            • $begingroup$
              No, this is like saying that the name of the function $cos$ is "trigonometric function". Exponential generating functions are just a type of generating function. These are a collection of methods for manipulating series by treating the terms of the series as coefficients in something like a Taylor series and then manipulating the function represented by that series, instead of the original series of numbers.
              $endgroup$
              – David Richerby
              13 hours ago








            • 1




              $begingroup$
              @DavidRicherby I guess you are right, but don't be too hard on JV. Recognizing this as an EGF could be useful in the process of identifying the function, just like noticing that something is a trigonometric function can be useful in precisely determining which function it is.
              $endgroup$
              – Tempestas Ludi
              12 hours ago














            • 1




              $begingroup$
              The formula in the question is with $k^n$, not $k_n$. And that in the paper is with $k^{(n)}$, which is quite different.
              $endgroup$
              – Yves Daoust
              14 hours ago






            • 1




              $begingroup$
              @YvesDaoust, you are right, I am just would have liked to help to celani99.
              $endgroup$
              – JV.Stalker
              14 hours ago










            • $begingroup$
              No, this is like saying that the name of the function $cos$ is "trigonometric function". Exponential generating functions are just a type of generating function. These are a collection of methods for manipulating series by treating the terms of the series as coefficients in something like a Taylor series and then manipulating the function represented by that series, instead of the original series of numbers.
              $endgroup$
              – David Richerby
              13 hours ago








            • 1




              $begingroup$
              @DavidRicherby I guess you are right, but don't be too hard on JV. Recognizing this as an EGF could be useful in the process of identifying the function, just like noticing that something is a trigonometric function can be useful in precisely determining which function it is.
              $endgroup$
              – Tempestas Ludi
              12 hours ago








            1




            1




            $begingroup$
            The formula in the question is with $k^n$, not $k_n$. And that in the paper is with $k^{(n)}$, which is quite different.
            $endgroup$
            – Yves Daoust
            14 hours ago




            $begingroup$
            The formula in the question is with $k^n$, not $k_n$. And that in the paper is with $k^{(n)}$, which is quite different.
            $endgroup$
            – Yves Daoust
            14 hours ago




            1




            1




            $begingroup$
            @YvesDaoust, you are right, I am just would have liked to help to celani99.
            $endgroup$
            – JV.Stalker
            14 hours ago




            $begingroup$
            @YvesDaoust, you are right, I am just would have liked to help to celani99.
            $endgroup$
            – JV.Stalker
            14 hours ago












            $begingroup$
            No, this is like saying that the name of the function $cos$ is "trigonometric function". Exponential generating functions are just a type of generating function. These are a collection of methods for manipulating series by treating the terms of the series as coefficients in something like a Taylor series and then manipulating the function represented by that series, instead of the original series of numbers.
            $endgroup$
            – David Richerby
            13 hours ago






            $begingroup$
            No, this is like saying that the name of the function $cos$ is "trigonometric function". Exponential generating functions are just a type of generating function. These are a collection of methods for manipulating series by treating the terms of the series as coefficients in something like a Taylor series and then manipulating the function represented by that series, instead of the original series of numbers.
            $endgroup$
            – David Richerby
            13 hours ago






            1




            1




            $begingroup$
            @DavidRicherby I guess you are right, but don't be too hard on JV. Recognizing this as an EGF could be useful in the process of identifying the function, just like noticing that something is a trigonometric function can be useful in precisely determining which function it is.
            $endgroup$
            – Tempestas Ludi
            12 hours ago




            $begingroup$
            @DavidRicherby I guess you are right, but don't be too hard on JV. Recognizing this as an EGF could be useful in the process of identifying the function, just like noticing that something is a trigonometric function can be useful in precisely determining which function it is.
            $endgroup$
            – Tempestas Ludi
            12 hours ago










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