Is there a name for this series?
$begingroup$
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
New contributor
$endgroup$
|
show 1 more 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
New contributor
$endgroup$
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
|
show 1 more 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
New contributor
$endgroup$
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
New contributor
New contributor
edited 14 hours ago
celani99
New contributor
asked 15 hours ago
celani99celani99
193
193
New contributor
New contributor
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
|
show 1 more comment
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
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
$$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$
This can be rewritten as
$$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.
$endgroup$
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.
$endgroup$
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
add a comment |
$begingroup$
Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.
$sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$
$endgroup$
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
add a comment |
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3 Answers
3
active
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3 Answers
3
active
oldest
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active
oldest
votes
$begingroup$
$$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$
This can be rewritten as
$$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.
$endgroup$
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.
$endgroup$
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.
$endgroup$
$$sum_{n=1}^inftyfrac{(-ik)^n}{(-i),n!}=i(e^{-ik}-1)=i(cos k-1-isin k).$$
This can be rewritten as
$$2icosfrac k2left(cosfrac k2-isinfrac k2right)=2icosfrac k2e^{-ik/2}$$ but there is little benefit.
edited 14 hours ago
answered 15 hours ago
Yves DaoustYves Daoust
128k675227
128k675227
add a comment |
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.
$endgroup$
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
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.
$endgroup$
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
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.
$endgroup$
$$sumlimits_{n=1}^infty frac{(-i)^{n-1}}{n!}k^n=sumlimits_{n=1}^infty frac{(-i)(-i)^{n-1}}{(-i)n!}k^n=frac{1}{-i}sumlimits_{n=1}^inftyfrac{(-i)^n}{n!}k^n=i(e^{k}-1).$$ So I would say yes it is known and there is a name for it -- exponential function.
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
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
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
add a comment |
$begingroup$
Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.
$sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$
$endgroup$
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
add a comment |
$begingroup$
Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.
$sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$
$endgroup$
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
add a comment |
$begingroup$
Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.
$sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$
$endgroup$
Think the name of the function is "Exponential generating function (EGF)" see Wikipedia.
$sumlimits_{n=1}^infty ik_n frac{(-i)^n}{n!}=EGF(ik_n,-i)-ik(0)$
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
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
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.
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– Tempestas Ludi
12 hours ago
1
1
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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.
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– Yves Daoust
14 hours ago
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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.
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– Yves Daoust
14 hours ago
1
1
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@YvesDaoust, you are right, I am just would have liked to help to celani99.
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– JV.Stalker
14 hours ago
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@YvesDaoust, you are right, I am just would have liked to help to celani99.
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– JV.Stalker
14 hours ago
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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.
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– David Richerby
13 hours ago
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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.
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– 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
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
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After checking the paper, the formula is much more complicated than that. Voting to close.
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– Yves Daoust
15 hours ago
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@YvesDaoust Should I delete the question instead?
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– celani99
14 hours ago
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You still have the option of updating the question to ask for clarification about the formula.
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– Yves Daoust
14 hours ago
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@YvesDaoust Is that edit sufficient, or should I add any other information?
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– celani99
14 hours ago
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$^{(n)}$, not $^n$ !
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– Yves Daoust
13 hours ago