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$begingroup$
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
asked 1 hour ago
celani99celani99
213
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.
$endgroup$
add a comment |
$begingroup$
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
asked 1 hour ago
celani99celani99
213
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
1 hour ago
$begingroup$
@YvesDaoust Should I delete the question instead?
$endgroup$
– celani99
32 mins ago
$begingroup$
You still have the option of updating the question to ask for clarification about the formula.
$endgroup$
– Yves Daoust
30 mins ago
$begingroup$
@YvesDaoust Is that edit sufficient, or should I add any other information?
$endgroup$
– celani99
15 mins ago
add a comment |
$begingroup$
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
asked 1 hour ago
celani99celani99
213
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
asked 1 hour ago
celani99celani99
213
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 1 hour ago
celani99celani99
213
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.
edited 16 mins ago
celani99
asked 1 hour ago
celani99celani99
213
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 1 hour ago
celani99celani99
213
asked 1 hour ago
celani99celani99
213
213
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
1 hour ago
$begingroup$
@YvesDaoust Should I delete the question instead?
$endgroup$
– celani99
32 mins ago
$begingroup$
You still have the option of updating the question to ask for clarification about the formula.
$endgroup$
– Yves Daoust
30 mins ago
$begingroup$
@YvesDaoust Is that edit sufficient, or should I add any other information?
$endgroup$
– celani99
15 mins ago
add a comment |
1
$begingroup$
After checking the paper, the formula is much more complicated than that. Voting to close.
$endgroup$
– Yves Daoust
1 hour ago
$begingroup$
@YvesDaoust Should I delete the question instead?
$endgroup$
– celani99
32 mins ago
$begingroup$
You still have the option of updating the question to ask for clarification about the formula.
$endgroup$
– Yves Daoust
30 mins ago
$begingroup$
@YvesDaoust Is that edit sufficient, or should I add any other information?
$endgroup$
– celani99
15 mins ago
1
1
$begingroup$
After checking the paper, the formula is much more complicated than that. Voting to close.
$endgroup$
– Yves Daoust
1 hour ago
$begingroup$
After checking the paper, the formula is much more complicated than that. Voting to close.
$endgroup$
– Yves Daoust
1 hour ago
$begingroup$
@YvesDaoust Should I delete the question instead?
$endgroup$
– celani99
32 mins ago
$begingroup$
@YvesDaoust Should I delete the question instead?
$endgroup$
– celani99
32 mins ago
$begingroup$
You still have the option of updating the question to ask for clarification about the formula.
$endgroup$
– Yves Daoust
30 mins ago
$begingroup$
You still have the option of updating the question to ask for clarification about the formula.
$endgroup$
– Yves Daoust
30 mins ago
$begingroup$
@YvesDaoust Is that edit sufficient, or should I add any other information?
$endgroup$
– celani99
15 mins ago
$begingroup$
@YvesDaoust Is that edit sufficient, or should I add any other information?
$endgroup$
– celani99
15 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{k}-1).$$ So I would say yes it is known and there is a name for it -- exponential function.
answered 1 hour ago
JamesJames
1,622217
$endgroup$
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
1 hour 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
47 mins ago
$begingroup$
@JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
$endgroup$
– celani99
38 mins ago
add a comment |
$begingroup$
$$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$
This can be rewritten as
$$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.
answered 1 hour ago
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.
$sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$
answered 35 mins ago
JV.StalkerJV.Stalker
84849
$endgroup$
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
29 mins ago
1
$begingroup$
@YvesDaoust, you are right, I am just would have liked to help to celani99.
$endgroup$
– JV.Stalker
22 mins ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{k}-1).$$ So I would say yes it is known and there is a name for it -- exponential function.
answered 1 hour ago
JamesJames
1,622217
$endgroup$
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
1 hour 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
47 mins ago
$begingroup$
@JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
$endgroup$
– celani99
38 mins ago
add a comment |
$begingroup$
$$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{k}-1).$$ So I would say yes it is known and there is a name for it -- exponential function.
answered 1 hour ago
JamesJames
1,622217
$endgroup$
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
1 hour 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
47 mins ago
$begingroup$
@JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
$endgroup$
– celani99
38 mins ago
add a comment |
$begingroup$
$$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{k}-1).$$ So I would say yes it is known and there is a name for it -- exponential function.
answered 1 hour ago
JamesJames
1,622217
$endgroup$
$$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{k}-1).$$ So I would say yes it is known and there is a name for it -- exponential function.
answered 1 hour ago
JamesJames
1,622217
edited 1 hour ago
answered 1 hour ago
JamesJames
1,622217
answered 1 hour ago
JamesJames
1,622217
answered 1 hour ago
JamesJames
1,622217
1,622217
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
1 hour 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
47 mins ago
$begingroup$
@JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
$endgroup$
– celani99
38 mins ago
add a comment |
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
1 hour 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
47 mins ago
$begingroup$
@JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
$endgroup$
– celani99
38 mins 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
1 hour 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
1 hour 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
47 mins 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
47 mins ago
$begingroup$
@JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
$endgroup$
– celani99
38 mins ago
$begingroup$
@JV.Stalker, I think you are correct. The series I posted should be $k^{(n)}$, sorry.
$endgroup$
– celani99
38 mins ago
add a comment |
$begingroup$
$$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$
This can be rewritten as
$$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.
answered 1 hour ago
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
$$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$
This can be rewritten as
$$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.
answered 1 hour ago
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
$$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$
This can be rewritten as
$$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.
answered 1 hour ago
Yves DaoustYves Daoust
128k675227
$endgroup$
$$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$
This can be rewritten as
$$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.
answered 1 hour ago
Yves DaoustYves Daoust
128k675227
edited 28 mins ago
answered 1 hour ago
Yves DaoustYves Daoust
128k675227
answered 1 hour ago
Yves DaoustYves Daoust
128k675227
answered 1 hour ago
Yves DaoustYves Daoust
128k675227
128k675227
add a comment |
add a comment |
$begingroup$
Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.
$sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$
answered 35 mins ago
JV.StalkerJV.Stalker
84849
$endgroup$
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
29 mins ago
1
$begingroup$
@YvesDaoust, you are right, I am just would have liked to help to celani99.
$endgroup$
– JV.Stalker
22 mins ago
add a comment |
$begingroup$
Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.
$sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$
answered 35 mins ago
JV.StalkerJV.Stalker
84849
$endgroup$
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
29 mins ago
1
$begingroup$
@YvesDaoust, you are right, I am just would have liked to help to celani99.
$endgroup$
– JV.Stalker
22 mins ago
add a comment |
$begingroup$
Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.
$sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$
answered 35 mins ago
JV.StalkerJV.Stalker
84849
$endgroup$
Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.
$sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$
answered 35 mins ago
JV.StalkerJV.Stalker
84849
edited 29 mins ago
answered 35 mins ago
JV.StalkerJV.Stalker
84849
answered 35 mins ago
JV.StalkerJV.Stalker
84849
answered 35 mins ago
JV.StalkerJV.Stalker
84849
84849
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
29 mins ago
1
$begingroup$
@YvesDaoust, you are right, I am just would have liked to help to celani99.
$endgroup$
– JV.Stalker
22 mins ago
add a comment |
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
29 mins ago
1
$begingroup$
@YvesDaoust, you are right, I am just would have liked to help to celani99.
$endgroup$
– JV.Stalker
22 mins 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
29 mins 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
29 mins ago
1
1
$begingroup$
@YvesDaoust, you are right, I am just would have liked to help to celani99.
$endgroup$
– JV.Stalker
22 mins ago
$begingroup$
@YvesDaoust, you are right, I am just would have liked to help to celani99.
$endgroup$
– JV.Stalker
22 mins ago
add a comment |
celani99 is a new contributor. Be nice, and check out our Code of Conduct.
celani99 is a new contributor. Be nice, and check out our Code of Conduct.
celani99 is a new contributor. Be nice, and check out our Code of Conduct.
celani99 is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
After checking the paper, the formula is much more complicated than that. Voting to close.
$endgroup$
– Yves Daoust
1 hour ago
$begingroup$
@YvesDaoust Should I delete the question instead?
$endgroup$
– celani99
32 mins ago
$begingroup$
You still have the option of updating the question to ask for clarification about the formula.
$endgroup$
– Yves Daoust
30 mins ago
$begingroup$
@YvesDaoust Is that edit sufficient, or should I add any other information?
$endgroup$
– celani99
15 mins ago