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How to extract coefficients of a generating function like this one, using a computer?
Finding all length-n words on an alphabet that have a specified number of each letterRandom filling of L-length line with l-length segmentsGenerating all “from n choose k” configurations of a simple listDefining Associated Stirling Numbers of the Second KindFind exponential generating function from the first few termsHow many equation systems can I build from five equations?How do I calculate the number of possible cases of polynomial equation (combinations) in Mathematica?Generating invertible matrix with lines within a given setCounting the permuted partitionsUsing FindInstance for Identifying Feasible Planar Solutions in a 3D Latin Hypercube
$begingroup$
For example if we have the generating function $G (x) = (1 + x + ... + x^k)^10$ and we want to calculate the coefficient of $x^3k$ as a function of $k $:
What is the best way to go about it using Mathematica?
combinatorics
asked 8 hours ago
oxynoia oxynoia
61 bronze badge
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oxynoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
For example if we have the generating function $G (x) = (1 + x + ... + x^k)^10$ and we want to calculate the coefficient of $x^3k$ as a function of $k $:
What is the best way to go about it using Mathematica?
combinatorics
asked 8 hours ago
oxynoia oxynoia
61 bronze badge
New contributor
oxynoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Assuming[k [Element] Integers && k > 1, SeriesCoefficient[(-1 + x^(1 + k))^10/(-1 + x)^10, x, 0, 3 k] // FullSimplify].
$endgroup$
– AccidentalFourierTransform
6 hours ago
add a comment |
$begingroup$
For example if we have the generating function $G (x) = (1 + x + ... + x^k)^10$ and we want to calculate the coefficient of $x^3k$ as a function of $k $:
What is the best way to go about it using Mathematica?
combinatorics
asked 8 hours ago
oxynoia oxynoia
61 bronze badge
New contributor
oxynoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
For example if we have the generating function $G (x) = (1 + x + ... + x^k)^10$ and we want to calculate the coefficient of $x^3k$ as a function of $k $:
What is the best way to go about it using Mathematica?
combinatorics
combinatorics
asked 8 hours ago
oxynoia oxynoia
61 bronze badge
New contributor
oxynoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
oxynoia oxynoia
61 bronze badge
New contributor
oxynoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
oxynoia oxynoia
61 bronze badge
New contributor
oxynoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
oxynoia oxynoia
61 bronze badge
asked 8 hours ago
oxynoia oxynoia
61 bronze badge
61 bronze badge
New contributor
oxynoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
oxynoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Assuming[k [Element] Integers && k > 1, SeriesCoefficient[(-1 + x^(1 + k))^10/(-1 + x)^10, x, 0, 3 k] // FullSimplify].
$endgroup$
– AccidentalFourierTransform
6 hours ago
add a comment |
1
$begingroup$
Assuming[k [Element] Integers && k > 1, SeriesCoefficient[(-1 + x^(1 + k))^10/(-1 + x)^10, x, 0, 3 k] // FullSimplify].
$endgroup$
– AccidentalFourierTransform
6 hours ago
1
1
$begingroup$
Assuming[k [Element] Integers && k > 1, SeriesCoefficient[(-1 + x^(1 + k))^10/(-1 + x)^10, x, 0, 3 k] // FullSimplify].$endgroup$
– AccidentalFourierTransform
6 hours ago
$begingroup$
Assuming[k [Element] Integers && k > 1, SeriesCoefficient[(-1 + x^(1 + k))^10/(-1 + x)^10, x, 0, 3 k] // FullSimplify].$endgroup$
– AccidentalFourierTransform
6 hours ago
add a comment |
2 Answers
2
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$begingroup$
Clear["Global`*"]
The closed-form of the Sum is
G[x_, k_] = Sum[x^n, n, 0, k]^10
(* (-1 + x^(1 + k))^10/(-1 + x)^10 *)
The coefficient for the x^(3k) term of G[x, k] is
coef3k[k_] =
FindSequenceFunction[
Table[SeriesCoefficient[G[x, k], x, 0, 3 k], k, 1, 15], k] //
FullSimplify
(* (1/90720)(1 + k) (2 + k) (3 + k) (15120 +
k (57552 + k (121438 + k (137565 + k (89110 + k (29163 + 3652 k)))))) *)
Verifying the result for k in the interval 0, 200
And @@ Table[
coef3k[k] == SeriesCoefficient[G[x, k], x, 0, 3 k],
k, 0, 200]
(* True *)
answered 7 hours ago
Bob HanlonBob Hanlon
63.3k3 gold badges36 silver badges99 bronze badges
$endgroup$
add a comment |
$begingroup$
Define the function
a[n_, k_] := SeriesCoefficient[ Sum[x^i, i, 0, k]^10, x, 0, n];
and you want a[3 k, k]. By the way, this is a $9$th degree polynomial function as given by
FindSequenceFunction[Table[a[3 k, k], k, 11]]
answered 7 hours ago
SomosSomos
2,7801 gold badge2 silver badges11 bronze badges
$endgroup$
add a comment |
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2 Answers
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oldest
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2 Answers
2
active
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active
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active
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votes
$begingroup$
Clear["Global`*"]
The closed-form of the Sum is
G[x_, k_] = Sum[x^n, n, 0, k]^10
(* (-1 + x^(1 + k))^10/(-1 + x)^10 *)
The coefficient for the x^(3k) term of G[x, k] is
coef3k[k_] =
FindSequenceFunction[
Table[SeriesCoefficient[G[x, k], x, 0, 3 k], k, 1, 15], k] //
FullSimplify
(* (1/90720)(1 + k) (2 + k) (3 + k) (15120 +
k (57552 + k (121438 + k (137565 + k (89110 + k (29163 + 3652 k)))))) *)
Verifying the result for k in the interval 0, 200
And @@ Table[
coef3k[k] == SeriesCoefficient[G[x, k], x, 0, 3 k],
k, 0, 200]
(* True *)
answered 7 hours ago
Bob HanlonBob Hanlon
63.3k3 gold badges36 silver badges99 bronze badges
$endgroup$
add a comment |
$begingroup$
Clear["Global`*"]
The closed-form of the Sum is
G[x_, k_] = Sum[x^n, n, 0, k]^10
(* (-1 + x^(1 + k))^10/(-1 + x)^10 *)
The coefficient for the x^(3k) term of G[x, k] is
coef3k[k_] =
FindSequenceFunction[
Table[SeriesCoefficient[G[x, k], x, 0, 3 k], k, 1, 15], k] //
FullSimplify
(* (1/90720)(1 + k) (2 + k) (3 + k) (15120 +
k (57552 + k (121438 + k (137565 + k (89110 + k (29163 + 3652 k)))))) *)
Verifying the result for k in the interval 0, 200
And @@ Table[
coef3k[k] == SeriesCoefficient[G[x, k], x, 0, 3 k],
k, 0, 200]
(* True *)
answered 7 hours ago
Bob HanlonBob Hanlon
63.3k3 gold badges36 silver badges99 bronze badges
$endgroup$
add a comment |
$begingroup$
Clear["Global`*"]
The closed-form of the Sum is
G[x_, k_] = Sum[x^n, n, 0, k]^10
(* (-1 + x^(1 + k))^10/(-1 + x)^10 *)
The coefficient for the x^(3k) term of G[x, k] is
coef3k[k_] =
FindSequenceFunction[
Table[SeriesCoefficient[G[x, k], x, 0, 3 k], k, 1, 15], k] //
FullSimplify
(* (1/90720)(1 + k) (2 + k) (3 + k) (15120 +
k (57552 + k (121438 + k (137565 + k (89110 + k (29163 + 3652 k)))))) *)
Verifying the result for k in the interval 0, 200
And @@ Table[
coef3k[k] == SeriesCoefficient[G[x, k], x, 0, 3 k],
k, 0, 200]
(* True *)
answered 7 hours ago
Bob HanlonBob Hanlon
63.3k3 gold badges36 silver badges99 bronze badges
$endgroup$
Clear["Global`*"]
The closed-form of the Sum is
G[x_, k_] = Sum[x^n, n, 0, k]^10
(* (-1 + x^(1 + k))^10/(-1 + x)^10 *)
The coefficient for the x^(3k) term of G[x, k] is
coef3k[k_] =
FindSequenceFunction[
Table[SeriesCoefficient[G[x, k], x, 0, 3 k], k, 1, 15], k] //
FullSimplify
(* (1/90720)(1 + k) (2 + k) (3 + k) (15120 +
k (57552 + k (121438 + k (137565 + k (89110 + k (29163 + 3652 k)))))) *)
Verifying the result for k in the interval 0, 200
And @@ Table[
coef3k[k] == SeriesCoefficient[G[x, k], x, 0, 3 k],
k, 0, 200]
(* True *)
answered 7 hours ago
Bob HanlonBob Hanlon
63.3k3 gold badges36 silver badges99 bronze badges
answered 7 hours ago
Bob HanlonBob Hanlon
63.3k3 gold badges36 silver badges99 bronze badges
answered 7 hours ago
Bob HanlonBob Hanlon
63.3k3 gold badges36 silver badges99 bronze badges
answered 7 hours ago
Bob HanlonBob Hanlon
63.3k3 gold badges36 silver badges99 bronze badges
63.3k3 gold badges36 silver badges99 bronze badges
add a comment |
add a comment |
$begingroup$
Define the function
a[n_, k_] := SeriesCoefficient[ Sum[x^i, i, 0, k]^10, x, 0, n];
and you want a[3 k, k]. By the way, this is a $9$th degree polynomial function as given by
FindSequenceFunction[Table[a[3 k, k], k, 11]]
answered 7 hours ago
SomosSomos
2,7801 gold badge2 silver badges11 bronze badges
$endgroup$
add a comment |
$begingroup$
Define the function
a[n_, k_] := SeriesCoefficient[ Sum[x^i, i, 0, k]^10, x, 0, n];
and you want a[3 k, k]. By the way, this is a $9$th degree polynomial function as given by
FindSequenceFunction[Table[a[3 k, k], k, 11]]
answered 7 hours ago
SomosSomos
2,7801 gold badge2 silver badges11 bronze badges
$endgroup$
add a comment |
$begingroup$
Define the function
a[n_, k_] := SeriesCoefficient[ Sum[x^i, i, 0, k]^10, x, 0, n];
and you want a[3 k, k]. By the way, this is a $9$th degree polynomial function as given by
FindSequenceFunction[Table[a[3 k, k], k, 11]]
answered 7 hours ago
SomosSomos
2,7801 gold badge2 silver badges11 bronze badges
$endgroup$
Define the function
a[n_, k_] := SeriesCoefficient[ Sum[x^i, i, 0, k]^10, x, 0, n];
and you want a[3 k, k]. By the way, this is a $9$th degree polynomial function as given by
FindSequenceFunction[Table[a[3 k, k], k, 11]]
answered 7 hours ago
SomosSomos
2,7801 gold badge2 silver badges11 bronze badges
edited 7 hours ago
answered 7 hours ago
SomosSomos
2,7801 gold badge2 silver badges11 bronze badges
answered 7 hours ago
SomosSomos
2,7801 gold badge2 silver badges11 bronze badges
answered 7 hours ago
SomosSomos
2,7801 gold badge2 silver badges11 bronze badges
2,7801 gold badge2 silver badges11 bronze badges
add a comment |
add a comment |
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1
$begingroup$
Assuming[k [Element] Integers && k > 1, SeriesCoefficient[(-1 + x^(1 + k))^10/(-1 + x)^10, x, 0, 3 k] // FullSimplify].$endgroup$
– AccidentalFourierTransform
6 hours ago