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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

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

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

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    $\begingroup$ 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
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$$\sum_{n=1}^\infty\frac{(-ik)^n}{(-i)\,n!}=i(e^{-ik}-1)=i(\cos k-1-i\sin k).$$


This can be rewritten as

$$2i\cos\frac k2\left(\cos\frac k2-i\sin\frac k2\right)=2i\cos\frac k2e^{-ik/2}$$ but there is little benefit.

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

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

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

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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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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