General formula for Differentiation Operator
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
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I was considering the Operator
$$
x,fracrm drm dx
$$
and applying it $n$ times to an arbitrary function $f(x)$.
Is there a general formula for it?
I started with the first few
beginalign
left(x,fracrm drm dxright)^1 f(x) &= x f' \
left(x,fracrm drm dxright)^2 f(x) &= x f' + x^2 f'' \
left(x,fracrm drm dxright)^3 f(x) &= x f' + 3x^2 f'' + x^3 f''' \
left(x,fracrm drm dxright)^4 f(x) &= x f' + 7 x^2 f'' + 6x^3 f''' + x^4 f'''' \
vdots
endalign
but I don't see any pattern yet.
Obviously the first and last coefficients are always $1$.
I figured if I start with any $n$ of the form
$$
left(x,fracrm drm dxright)^n f(x) = sum_k=1^n a_k , x^k f^(k)(x) tag1
$$
where $a_1=a_n=1$, then recursively
beginalign
left(x,fracrm drm dxright)^n+1 f(x) &= xf^(1)(x) + sum_k=2^n left(k , a_k + a_k-1right) x^k f^(k)(x) + x^n+1 f^(n+1)(x) \
&= sum_k=1^n+1 left(k , a_k + a_k-1right) x^k f^(k)(x) tag2
endalign
where $a_0=a_n+1=0$.
From (1) and (2) we get for fixed $k$ the coupled recurrence
$$
a_k(n+1) = k, a_k(n) + a_k-1(n)
$$
which can be put in matrix form
beginalign
beginpmatrix a_1(n+1) \ a_2(n+1) \ a_3(n+1) \ vdots \ a_n-1(n+1) \ a_n(n+1) endpmatrix &= beginpmatrix 1 & 0 & 0 & dots & 0 & 0 \ 1 & 2 & 0 & dots & 0 & 0 \ 0 & 1 & 3 & dots & 0 & 0 \ vdots & vdots & vdots & dots & vdots & vdots \ 0 & 0 & 0 & dots & n-1 & 0 \ 0 & 0 & 0 & dots & 1 & n endpmatrix beginpmatrix a_1(n) \ a_2(n) \ a_3(n) \ vdots \ a_n-1(n) \ a_n(n) endpmatrix \
&=beginpmatrix 1 & 0 & 0 & dots & 0 & 0 \ 1 & 2 & 0 & dots & 0 & 0 \ 0 & 1 & 3 & dots & 0 & 0 \ vdots & vdots & vdots & dots & vdots & vdots \ 0 & 0 & 0 & dots & n-1 & 0 \ 0 & 0 & 0 & dots & 1 & n endpmatrix^n beginpmatrix 1 \ 0 \ 0 \ vdots \ 0 \ 0 endpmatrix , .
endalign
So either I'm too stupid or there is no obvious one.
derivatives summation recurrence-relations differential-operators
asked Sep 4 at 12:47
Diger
77229
add a comment |Â
up vote
4
down vote
favorite
I was considering the Operator
$$
x,fracrm drm dx
$$
and applying it $n$ times to an arbitrary function $f(x)$.
Is there a general formula for it?
I started with the first few
beginalign
left(x,fracrm drm dxright)^1 f(x) &= x f' \
left(x,fracrm drm dxright)^2 f(x) &= x f' + x^2 f'' \
left(x,fracrm drm dxright)^3 f(x) &= x f' + 3x^2 f'' + x^3 f''' \
left(x,fracrm drm dxright)^4 f(x) &= x f' + 7 x^2 f'' + 6x^3 f''' + x^4 f'''' \
vdots
endalign
but I don't see any pattern yet.
Obviously the first and last coefficients are always $1$.
I figured if I start with any $n$ of the form
$$
left(x,fracrm drm dxright)^n f(x) = sum_k=1^n a_k , x^k f^(k)(x) tag1
$$
where $a_1=a_n=1$, then recursively
beginalign
left(x,fracrm drm dxright)^n+1 f(x) &= xf^(1)(x) + sum_k=2^n left(k , a_k + a_k-1right) x^k f^(k)(x) + x^n+1 f^(n+1)(x) \
&= sum_k=1^n+1 left(k , a_k + a_k-1right) x^k f^(k)(x) tag2
endalign
where $a_0=a_n+1=0$.
From (1) and (2) we get for fixed $k$ the coupled recurrence
$$
a_k(n+1) = k, a_k(n) + a_k-1(n)
$$
which can be put in matrix form
beginalign
beginpmatrix a_1(n+1) \ a_2(n+1) \ a_3(n+1) \ vdots \ a_n-1(n+1) \ a_n(n+1) endpmatrix &= beginpmatrix 1 & 0 & 0 & dots & 0 & 0 \ 1 & 2 & 0 & dots & 0 & 0 \ 0 & 1 & 3 & dots & 0 & 0 \ vdots & vdots & vdots & dots & vdots & vdots \ 0 & 0 & 0 & dots & n-1 & 0 \ 0 & 0 & 0 & dots & 1 & n endpmatrix beginpmatrix a_1(n) \ a_2(n) \ a_3(n) \ vdots \ a_n-1(n) \ a_n(n) endpmatrix \
&=beginpmatrix 1 & 0 & 0 & dots & 0 & 0 \ 1 & 2 & 0 & dots & 0 & 0 \ 0 & 1 & 3 & dots & 0 & 0 \ vdots & vdots & vdots & dots & vdots & vdots \ 0 & 0 & 0 & dots & n-1 & 0 \ 0 & 0 & 0 & dots & 1 & n endpmatrix^n beginpmatrix 1 \ 0 \ 0 \ vdots \ 0 \ 0 endpmatrix , .
endalign
So either I'm too stupid or there is no obvious one.
derivatives summation recurrence-relations differential-operators
asked Sep 4 at 12:47
Diger
77229
This might be relevant to your interests: Weyl Algebra. I once had to do something like this, and I was just as confused as you are. However, in my problem I only needed to know information about the first and last terms. You can make inductive arguments about those, I think.
– Prototank
Sep 4 at 12:57
Well the first and last terms are always one, so there is not much to do or?
– Diger
Sep 4 at 13:01
I guess my question is: what do you need the general form for? It is fine if you are just curious, I was assuming you were trying to show something else and that this is a piece of the puzzle.
– Prototank
Sep 4 at 13:01
@Diger, I don't think you're stupid. However you transformed your problem into a boundary problem, if you can solve that you can find a closed form solution.
– user8469759
Sep 4 at 13:02
1
I want to know the growth of the coefficients in order to estimate a coefficient of a series. I need radius of convergence.
– Diger
Sep 4 at 13:03
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I was considering the Operator
$$
x,fracrm drm dx
$$
and applying it $n$ times to an arbitrary function $f(x)$.
Is there a general formula for it?
I started with the first few
beginalign
left(x,fracrm drm dxright)^1 f(x) &= x f' \
left(x,fracrm drm dxright)^2 f(x) &= x f' + x^2 f'' \
left(x,fracrm drm dxright)^3 f(x) &= x f' + 3x^2 f'' + x^3 f''' \
left(x,fracrm drm dxright)^4 f(x) &= x f' + 7 x^2 f'' + 6x^3 f''' + x^4 f'''' \
vdots
endalign
but I don't see any pattern yet.
Obviously the first and last coefficients are always $1$.
I figured if I start with any $n$ of the form
$$
left(x,fracrm drm dxright)^n f(x) = sum_k=1^n a_k , x^k f^(k)(x) tag1
$$
where $a_1=a_n=1$, then recursively
beginalign
left(x,fracrm drm dxright)^n+1 f(x) &= xf^(1)(x) + sum_k=2^n left(k , a_k + a_k-1right) x^k f^(k)(x) + x^n+1 f^(n+1)(x) \
&= sum_k=1^n+1 left(k , a_k + a_k-1right) x^k f^(k)(x) tag2
endalign
where $a_0=a_n+1=0$.
From (1) and (2) we get for fixed $k$ the coupled recurrence
$$
a_k(n+1) = k, a_k(n) + a_k-1(n)
$$
which can be put in matrix form
beginalign
beginpmatrix a_1(n+1) \ a_2(n+1) \ a_3(n+1) \ vdots \ a_n-1(n+1) \ a_n(n+1) endpmatrix &= beginpmatrix 1 & 0 & 0 & dots & 0 & 0 \ 1 & 2 & 0 & dots & 0 & 0 \ 0 & 1 & 3 & dots & 0 & 0 \ vdots & vdots & vdots & dots & vdots & vdots \ 0 & 0 & 0 & dots & n-1 & 0 \ 0 & 0 & 0 & dots & 1 & n endpmatrix beginpmatrix a_1(n) \ a_2(n) \ a_3(n) \ vdots \ a_n-1(n) \ a_n(n) endpmatrix \
&=beginpmatrix 1 & 0 & 0 & dots & 0 & 0 \ 1 & 2 & 0 & dots & 0 & 0 \ 0 & 1 & 3 & dots & 0 & 0 \ vdots & vdots & vdots & dots & vdots & vdots \ 0 & 0 & 0 & dots & n-1 & 0 \ 0 & 0 & 0 & dots & 1 & n endpmatrix^n beginpmatrix 1 \ 0 \ 0 \ vdots \ 0 \ 0 endpmatrix , .
endalign
So either I'm too stupid or there is no obvious one.
derivatives summation recurrence-relations differential-operators
asked Sep 4 at 12:47
Diger
77229
I was considering the Operator
$$
x,fracrm drm dx
$$
and applying it $n$ times to an arbitrary function $f(x)$.
Is there a general formula for it?
I started with the first few
beginalign
left(x,fracrm drm dxright)^1 f(x) &= x f' \
left(x,fracrm drm dxright)^2 f(x) &= x f' + x^2 f'' \
left(x,fracrm drm dxright)^3 f(x) &= x f' + 3x^2 f'' + x^3 f''' \
left(x,fracrm drm dxright)^4 f(x) &= x f' + 7 x^2 f'' + 6x^3 f''' + x^4 f'''' \
vdots
endalign
but I don't see any pattern yet.
Obviously the first and last coefficients are always $1$.
I figured if I start with any $n$ of the form
$$
left(x,fracrm drm dxright)^n f(x) = sum_k=1^n a_k , x^k f^(k)(x) tag1
$$
where $a_1=a_n=1$, then recursively
beginalign
left(x,fracrm drm dxright)^n+1 f(x) &= xf^(1)(x) + sum_k=2^n left(k , a_k + a_k-1right) x^k f^(k)(x) + x^n+1 f^(n+1)(x) \
&= sum_k=1^n+1 left(k , a_k + a_k-1right) x^k f^(k)(x) tag2
endalign
where $a_0=a_n+1=0$.
From (1) and (2) we get for fixed $k$ the coupled recurrence
$$
a_k(n+1) = k, a_k(n) + a_k-1(n)
$$
which can be put in matrix form
beginalign
beginpmatrix a_1(n+1) \ a_2(n+1) \ a_3(n+1) \ vdots \ a_n-1(n+1) \ a_n(n+1) endpmatrix &= beginpmatrix 1 & 0 & 0 & dots & 0 & 0 \ 1 & 2 & 0 & dots & 0 & 0 \ 0 & 1 & 3 & dots & 0 & 0 \ vdots & vdots & vdots & dots & vdots & vdots \ 0 & 0 & 0 & dots & n-1 & 0 \ 0 & 0 & 0 & dots & 1 & n endpmatrix beginpmatrix a_1(n) \ a_2(n) \ a_3(n) \ vdots \ a_n-1(n) \ a_n(n) endpmatrix \
&=beginpmatrix 1 & 0 & 0 & dots & 0 & 0 \ 1 & 2 & 0 & dots & 0 & 0 \ 0 & 1 & 3 & dots & 0 & 0 \ vdots & vdots & vdots & dots & vdots & vdots \ 0 & 0 & 0 & dots & n-1 & 0 \ 0 & 0 & 0 & dots & 1 & n endpmatrix^n beginpmatrix 1 \ 0 \ 0 \ vdots \ 0 \ 0 endpmatrix , .
endalign
So either I'm too stupid or there is no obvious one.
derivatives summation recurrence-relations differential-operators
derivatives summation recurrence-relations differential-operators
asked Sep 4 at 12:47
Diger
77229
asked Sep 4 at 12:47
Diger
77229
edited Sep 4 at 14:08
asked Sep 4 at 12:47
Diger
77229
asked Sep 4 at 12:47
Diger
77229
asked Sep 4 at 12:47
Diger
77229
77229
This might be relevant to your interests: Weyl Algebra. I once had to do something like this, and I was just as confused as you are. However, in my problem I only needed to know information about the first and last terms. You can make inductive arguments about those, I think.
– Prototank
Sep 4 at 12:57
Well the first and last terms are always one, so there is not much to do or?
– Diger
Sep 4 at 13:01
I guess my question is: what do you need the general form for? It is fine if you are just curious, I was assuming you were trying to show something else and that this is a piece of the puzzle.
– Prototank
Sep 4 at 13:01
@Diger, I don't think you're stupid. However you transformed your problem into a boundary problem, if you can solve that you can find a closed form solution.
– user8469759
Sep 4 at 13:02
1
I want to know the growth of the coefficients in order to estimate a coefficient of a series. I need radius of convergence.
– Diger
Sep 4 at 13:03
add a comment |Â
This might be relevant to your interests: Weyl Algebra. I once had to do something like this, and I was just as confused as you are. However, in my problem I only needed to know information about the first and last terms. You can make inductive arguments about those, I think.
– Prototank
Sep 4 at 12:57
Well the first and last terms are always one, so there is not much to do or?
– Diger
Sep 4 at 13:01
I guess my question is: what do you need the general form for? It is fine if you are just curious, I was assuming you were trying to show something else and that this is a piece of the puzzle.
– Prototank
Sep 4 at 13:01
@Diger, I don't think you're stupid. However you transformed your problem into a boundary problem, if you can solve that you can find a closed form solution.
– user8469759
Sep 4 at 13:02
1
I want to know the growth of the coefficients in order to estimate a coefficient of a series. I need radius of convergence.
– Diger
Sep 4 at 13:03
This might be relevant to your interests: Weyl Algebra. I once had to do something like this, and I was just as confused as you are. However, in my problem I only needed to know information about the first and last terms. You can make inductive arguments about those, I think.
– Prototank
Sep 4 at 12:57
This might be relevant to your interests: Weyl Algebra. I once had to do something like this, and I was just as confused as you are. However, in my problem I only needed to know information about the first and last terms. You can make inductive arguments about those, I think.
– Prototank
Sep 4 at 12:57
Well the first and last terms are always one, so there is not much to do or?
– Diger
Sep 4 at 13:01
Well the first and last terms are always one, so there is not much to do or?
– Diger
Sep 4 at 13:01
I guess my question is: what do you need the general form for? It is fine if you are just curious, I was assuming you were trying to show something else and that this is a piece of the puzzle.
– Prototank
Sep 4 at 13:01
I guess my question is: what do you need the general form for? It is fine if you are just curious, I was assuming you were trying to show something else and that this is a piece of the puzzle.
– Prototank
Sep 4 at 13:01
@Diger, I don't think you're stupid. However you transformed your problem into a boundary problem, if you can solve that you can find a closed form solution.
– user8469759
Sep 4 at 13:02
@Diger, I don't think you're stupid. However you transformed your problem into a boundary problem, if you can solve that you can find a closed form solution.
– user8469759
Sep 4 at 13:02
1
1
I want to know the growth of the coefficients in order to estimate a coefficient of a series. I need radius of convergence.
– Diger
Sep 4 at 13:03
I want to know the growth of the coefficients in order to estimate a coefficient of a series. I need radius of convergence.
– Diger
Sep 4 at 13:03
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The coefficients are the Stirling numbers of the second kind.
answered Sep 4 at 13:14
Hans Lundmark
33.6k564110
Must have been solved before ;) Thanks. How did you know?
– Diger
Sep 4 at 13:15
I recognized 1, 7, 6, 1. :-)
– Hans Lundmark
Sep 4 at 13:16
You could also compute a bit further and enter the coefficients of a later row into the OEIS.
– Hans Lundmark
Sep 4 at 13:19
Because you look at these numbers everyday? I would not even remember one of them...
– Diger
Sep 4 at 13:19
Thanks. Didn't know of this side. Very helpful.
– Diger
Sep 4 at 13:21
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The coefficients are the Stirling numbers of the second kind.
answered Sep 4 at 13:14
Hans Lundmark
33.6k564110
Must have been solved before ;) Thanks. How did you know?
– Diger
Sep 4 at 13:15
I recognized 1, 7, 6, 1. :-)
– Hans Lundmark
Sep 4 at 13:16
You could also compute a bit further and enter the coefficients of a later row into the OEIS.
– Hans Lundmark
Sep 4 at 13:19
Because you look at these numbers everyday? I would not even remember one of them...
– Diger
Sep 4 at 13:19
Thanks. Didn't know of this side. Very helpful.
– Diger
Sep 4 at 13:21
 |Â
show 1 more comment
up vote
3
down vote
accepted
The coefficients are the Stirling numbers of the second kind.
answered Sep 4 at 13:14
Hans Lundmark
33.6k564110
Must have been solved before ;) Thanks. How did you know?
– Diger
Sep 4 at 13:15
I recognized 1, 7, 6, 1. :-)
– Hans Lundmark
Sep 4 at 13:16
You could also compute a bit further and enter the coefficients of a later row into the OEIS.
– Hans Lundmark
Sep 4 at 13:19
Because you look at these numbers everyday? I would not even remember one of them...
– Diger
Sep 4 at 13:19
Thanks. Didn't know of this side. Very helpful.
– Diger
Sep 4 at 13:21
 |Â
show 1 more comment
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The coefficients are the Stirling numbers of the second kind.
answered Sep 4 at 13:14
Hans Lundmark
33.6k564110
The coefficients are the Stirling numbers of the second kind.
answered Sep 4 at 13:14
Hans Lundmark
33.6k564110
answered Sep 4 at 13:14
Hans Lundmark
33.6k564110
answered Sep 4 at 13:14
Hans Lundmark
33.6k564110
answered Sep 4 at 13:14
Hans Lundmark
33.6k564110
33.6k564110
Must have been solved before ;) Thanks. How did you know?
– Diger
Sep 4 at 13:15
I recognized 1, 7, 6, 1. :-)
– Hans Lundmark
Sep 4 at 13:16
You could also compute a bit further and enter the coefficients of a later row into the OEIS.
– Hans Lundmark
Sep 4 at 13:19
Because you look at these numbers everyday? I would not even remember one of them...
– Diger
Sep 4 at 13:19
Thanks. Didn't know of this side. Very helpful.
– Diger
Sep 4 at 13:21
 |Â
show 1 more comment
Must have been solved before ;) Thanks. How did you know?
– Diger
Sep 4 at 13:15
I recognized 1, 7, 6, 1. :-)
– Hans Lundmark
Sep 4 at 13:16
You could also compute a bit further and enter the coefficients of a later row into the OEIS.
– Hans Lundmark
Sep 4 at 13:19
Because you look at these numbers everyday? I would not even remember one of them...
– Diger
Sep 4 at 13:19
Thanks. Didn't know of this side. Very helpful.
– Diger
Sep 4 at 13:21
Must have been solved before ;) Thanks. How did you know?
– Diger
Sep 4 at 13:15
Must have been solved before ;) Thanks. How did you know?
– Diger
Sep 4 at 13:15
I recognized 1, 7, 6, 1. :-)
– Hans Lundmark
Sep 4 at 13:16
I recognized 1, 7, 6, 1. :-)
– Hans Lundmark
Sep 4 at 13:16
You could also compute a bit further and enter the coefficients of a later row into the OEIS.
– Hans Lundmark
Sep 4 at 13:19
You could also compute a bit further and enter the coefficients of a later row into the OEIS.
– Hans Lundmark
Sep 4 at 13:19
Because you look at these numbers everyday? I would not even remember one of them...
– Diger
Sep 4 at 13:19
Because you look at these numbers everyday? I would not even remember one of them...
– Diger
Sep 4 at 13:19
Thanks. Didn't know of this side. Very helpful.
– Diger
Sep 4 at 13:21
Thanks. Didn't know of this side. Very helpful.
– Diger
Sep 4 at 13:21
 |Â
show 1 more comment
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This might be relevant to your interests: Weyl Algebra. I once had to do something like this, and I was just as confused as you are. However, in my problem I only needed to know information about the first and last terms. You can make inductive arguments about those, I think.
– Prototank
Sep 4 at 12:57
Well the first and last terms are always one, so there is not much to do or?
– Diger
Sep 4 at 13:01
I guess my question is: what do you need the general form for? It is fine if you are just curious, I was assuming you were trying to show something else and that this is a piece of the puzzle.
– Prototank
Sep 4 at 13:01
@Diger, I don't think you're stupid. However you transformed your problem into a boundary problem, if you can solve that you can find a closed form solution.
– user8469759
Sep 4 at 13:02
1
I want to know the growth of the coefficients in order to estimate a coefficient of a series. I need radius of convergence.
– Diger
Sep 4 at 13:03