Is there a name for this series?












2












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    After checking the paper, the formula is much more complicated than that. Voting to close.
    $endgroup$
    – Yves Daoust
    Feb 22 at 14:19












  • $begingroup$
    @YvesDaoust Should I delete the question instead?
    $endgroup$
    – celani99
    Feb 22 at 14:54










  • $begingroup$
    You still have the option of updating the question to ask for clarification about the formula.
    $endgroup$
    – Yves Daoust
    Feb 22 at 14:56










  • $begingroup$
    @YvesDaoust Is that edit sufficient, or should I add any other information?
    $endgroup$
    – celani99
    Feb 22 at 15:11










  • $begingroup$
    $^{(n)}$, not $^n$ !
    $endgroup$
    – Yves Daoust
    Feb 22 at 16:25
















2












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    After checking the paper, the formula is much more complicated than that. Voting to close.
    $endgroup$
    – Yves Daoust
    Feb 22 at 14:19












  • $begingroup$
    @YvesDaoust Should I delete the question instead?
    $endgroup$
    – celani99
    Feb 22 at 14:54










  • $begingroup$
    You still have the option of updating the question to ask for clarification about the formula.
    $endgroup$
    – Yves Daoust
    Feb 22 at 14:56










  • $begingroup$
    @YvesDaoust Is that edit sufficient, or should I add any other information?
    $endgroup$
    – celani99
    Feb 22 at 15:11










  • $begingroup$
    $^{(n)}$, not $^n$ !
    $endgroup$
    – Yves Daoust
    Feb 22 at 16:25














2












2








2





$begingroup$


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











$endgroup$




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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 22 at 15:10







celani99

















asked Feb 22 at 13:52









celani99celani99

173




173








  • 1




    $begingroup$
    After checking the paper, the formula is much more complicated than that. Voting to close.
    $endgroup$
    – Yves Daoust
    Feb 22 at 14:19












  • $begingroup$
    @YvesDaoust Should I delete the question instead?
    $endgroup$
    – celani99
    Feb 22 at 14:54










  • $begingroup$
    You still have the option of updating the question to ask for clarification about the formula.
    $endgroup$
    – Yves Daoust
    Feb 22 at 14:56










  • $begingroup$
    @YvesDaoust Is that edit sufficient, or should I add any other information?
    $endgroup$
    – celani99
    Feb 22 at 15:11










  • $begingroup$
    $^{(n)}$, not $^n$ !
    $endgroup$
    – Yves Daoust
    Feb 22 at 16:25














  • 1




    $begingroup$
    After checking the paper, the formula is much more complicated than that. Voting to close.
    $endgroup$
    – Yves Daoust
    Feb 22 at 14:19












  • $begingroup$
    @YvesDaoust Should I delete the question instead?
    $endgroup$
    – celani99
    Feb 22 at 14:54










  • $begingroup$
    You still have the option of updating the question to ask for clarification about the formula.
    $endgroup$
    – Yves Daoust
    Feb 22 at 14:56










  • $begingroup$
    @YvesDaoust Is that edit sufficient, or should I add any other information?
    $endgroup$
    – celani99
    Feb 22 at 15:11










  • $begingroup$
    $^{(n)}$, not $^n$ !
    $endgroup$
    – Yves Daoust
    Feb 22 at 16:25








1




1




$begingroup$
After checking the paper, the formula is much more complicated than that. Voting to close.
$endgroup$
– Yves Daoust
Feb 22 at 14:19






$begingroup$
After checking the paper, the formula is much more complicated than that. Voting to close.
$endgroup$
– Yves Daoust
Feb 22 at 14:19














$begingroup$
@YvesDaoust Should I delete the question instead?
$endgroup$
– celani99
Feb 22 at 14:54




$begingroup$
@YvesDaoust Should I delete the question instead?
$endgroup$
– celani99
Feb 22 at 14:54












$begingroup$
You still have the option of updating the question to ask for clarification about the formula.
$endgroup$
– Yves Daoust
Feb 22 at 14:56




$begingroup$
You still have the option of updating the question to ask for clarification about the formula.
$endgroup$
– Yves Daoust
Feb 22 at 14:56












$begingroup$
@YvesDaoust Is that edit sufficient, or should I add any other information?
$endgroup$
– celani99
Feb 22 at 15:11




$begingroup$
@YvesDaoust Is that edit sufficient, or should I add any other information?
$endgroup$
– celani99
Feb 22 at 15:11












$begingroup$
$^{(n)}$, not $^n$ !
$endgroup$
– Yves Daoust
Feb 22 at 16:25




$begingroup$
$^{(n)}$, not $^n$ !
$endgroup$
– Yves Daoust
Feb 22 at 16:25










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











$endgroup$





















    5












    $begingroup$

    $$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{-ik}-1).$$ 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
      Feb 22 at 14:06










    • $begingroup$
      @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
      Feb 22 at 14:39










    • $begingroup$
      @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
      $endgroup$
      – celani99
      Feb 22 at 14:48








    • 4




      $begingroup$
      Isn't it $e^{-ik}$ rather than $e^k$?
      $endgroup$
      – Acccumulation
      Feb 22 at 15:44










    • $begingroup$
      @Acccumulation: amazing so many people upvoted without noticing.
      $endgroup$
      – Yves Daoust
      Feb 22 at 16:48



















    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
      Feb 22 at 14:57






    • 1




      $begingroup$
      @YvesDaoust, you are right, I am just would have liked to help to celani99.
      $endgroup$
      – JV.Stalker
      Feb 22 at 15:04










    • $begingroup$
      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
      Feb 22 at 16:01








    • 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
      Feb 22 at 16:48











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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 Feb 22 at 14:58

























        answered Feb 22 at 14:10









        Yves DaoustYves Daoust

        131k676229




        131k676229























            5












            $begingroup$

            $$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{-ik}-1).$$ 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
              Feb 22 at 14:06










            • $begingroup$
              @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
              Feb 22 at 14:39










            • $begingroup$
              @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
              $endgroup$
              – celani99
              Feb 22 at 14:48








            • 4




              $begingroup$
              Isn't it $e^{-ik}$ rather than $e^k$?
              $endgroup$
              – Acccumulation
              Feb 22 at 15:44










            • $begingroup$
              @Acccumulation: amazing so many people upvoted without noticing.
              $endgroup$
              – Yves Daoust
              Feb 22 at 16:48
















            5












            $begingroup$

            $$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{-ik}-1).$$ 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
              Feb 22 at 14:06










            • $begingroup$
              @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
              Feb 22 at 14:39










            • $begingroup$
              @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
              $endgroup$
              – celani99
              Feb 22 at 14:48








            • 4




              $begingroup$
              Isn't it $e^{-ik}$ rather than $e^k$?
              $endgroup$
              – Acccumulation
              Feb 22 at 15:44










            • $begingroup$
              @Acccumulation: amazing so many people upvoted without noticing.
              $endgroup$
              – Yves Daoust
              Feb 22 at 16:48














            5












            5








            5





            $begingroup$

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






            share|cite|improve this answer











            $endgroup$



            $$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{-ik}-1).$$ 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 Feb 23 at 8:43

























            answered Feb 22 at 13:58









            JamesJames

            2,113422




            2,113422








            • 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
              Feb 22 at 14:06










            • $begingroup$
              @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
              Feb 22 at 14:39










            • $begingroup$
              @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
              $endgroup$
              – celani99
              Feb 22 at 14:48








            • 4




              $begingroup$
              Isn't it $e^{-ik}$ rather than $e^k$?
              $endgroup$
              – Acccumulation
              Feb 22 at 15:44










            • $begingroup$
              @Acccumulation: amazing so many people upvoted without noticing.
              $endgroup$
              – Yves Daoust
              Feb 22 at 16:48














            • 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
              Feb 22 at 14:06










            • $begingroup$
              @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
              Feb 22 at 14:39










            • $begingroup$
              @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
              $endgroup$
              – celani99
              Feb 22 at 14:48








            • 4




              $begingroup$
              Isn't it $e^{-ik}$ rather than $e^k$?
              $endgroup$
              – Acccumulation
              Feb 22 at 15:44










            • $begingroup$
              @Acccumulation: amazing so many people upvoted without noticing.
              $endgroup$
              – Yves Daoust
              Feb 22 at 16:48








            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
            Feb 22 at 14:06




            $begingroup$
            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
            Feb 22 at 14:06












            $begingroup$
            @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
            Feb 22 at 14:39




            $begingroup$
            @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
            Feb 22 at 14:39












            $begingroup$
            @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
            $endgroup$
            – celani99
            Feb 22 at 14:48






            $begingroup$
            @JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
            $endgroup$
            – celani99
            Feb 22 at 14:48






            4




            4




            $begingroup$
            Isn't it $e^{-ik}$ rather than $e^k$?
            $endgroup$
            – Acccumulation
            Feb 22 at 15:44




            $begingroup$
            Isn't it $e^{-ik}$ rather than $e^k$?
            $endgroup$
            – Acccumulation
            Feb 22 at 15:44












            $begingroup$
            @Acccumulation: amazing so many people upvoted without noticing.
            $endgroup$
            – Yves Daoust
            Feb 22 at 16:48




            $begingroup$
            @Acccumulation: amazing so many people upvoted without noticing.
            $endgroup$
            – Yves Daoust
            Feb 22 at 16:48











            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
              Feb 22 at 14:57






            • 1




              $begingroup$
              @YvesDaoust, you are right, I am just would have liked to help to celani99.
              $endgroup$
              – JV.Stalker
              Feb 22 at 15:04










            • $begingroup$
              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
              Feb 22 at 16:01








            • 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
              Feb 22 at 16:48
















            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
              Feb 22 at 14:57






            • 1




              $begingroup$
              @YvesDaoust, you are right, I am just would have liked to help to celani99.
              $endgroup$
              – JV.Stalker
              Feb 22 at 15:04










            • $begingroup$
              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
              Feb 22 at 16:01








            • 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
              Feb 22 at 16:48














            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 Feb 22 at 14:57

























            answered Feb 22 at 14:51









            JV.StalkerJV.Stalker

            93349




            93349








            • 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
              Feb 22 at 14:57






            • 1




              $begingroup$
              @YvesDaoust, you are right, I am just would have liked to help to celani99.
              $endgroup$
              – JV.Stalker
              Feb 22 at 15:04










            • $begingroup$
              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
              Feb 22 at 16:01








            • 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
              Feb 22 at 16:48














            • 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
              Feb 22 at 14:57






            • 1




              $begingroup$
              @YvesDaoust, you are right, I am just would have liked to help to celani99.
              $endgroup$
              – JV.Stalker
              Feb 22 at 15:04










            • $begingroup$
              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
              Feb 22 at 16:01








            • 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
              Feb 22 at 16:48








            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
            Feb 22 at 14:57




            $begingroup$
            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
            Feb 22 at 14:57




            1




            1




            $begingroup$
            @YvesDaoust, you are right, I am just would have liked to help to celani99.
            $endgroup$
            – JV.Stalker
            Feb 22 at 15:04




            $begingroup$
            @YvesDaoust, you are right, I am just would have liked to help to celani99.
            $endgroup$
            – JV.Stalker
            Feb 22 at 15:04












            $begingroup$
            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
            Feb 22 at 16:01






            $begingroup$
            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
            Feb 22 at 16:01






            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
            Feb 22 at 16:48




            $begingroup$
            @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
            Feb 22 at 16:48


















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