Analytical form of the binomial moments
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I'm trying to find a general closed-form expression for the following sum:
$$m_p(N) = sum_{n=0}^{N} binom{N}{n} n^p$$
where $N,pge 0$ are integers.
So far I've been able to evaluate:
$$
begin{eqnarray*}
sum_n binom{N}{n} n^0 & = & 2^N\
sum_n binom{N}{n} n^1 & = & 2^{N - 1} N\
sum_n binom{N}{n} n^2 & = & 2^{N - 2} (N + 1) N\
sum_n binom{N}{n} n^3 & = & 2^{N - 3} (N + 3) N^2
end{eqnarray*}
$$
but I don't see a general pattern here.
Is there a general analytical expression?
Note that this is not a textbook exercise. I'm not sure if a solution exists.
combinatorics summation binomial-coefficients
asked Jan 17 at 16:07
beckobecko
2,43131943
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add a comment |
$begingroup$
I'm trying to find a general closed-form expression for the following sum:
$$m_p(N) = sum_{n=0}^{N} binom{N}{n} n^p$$
where $N,pge 0$ are integers.
So far I've been able to evaluate:
$$
begin{eqnarray*}
sum_n binom{N}{n} n^0 & = & 2^N\
sum_n binom{N}{n} n^1 & = & 2^{N - 1} N\
sum_n binom{N}{n} n^2 & = & 2^{N - 2} (N + 1) N\
sum_n binom{N}{n} n^3 & = & 2^{N - 3} (N + 3) N^2
end{eqnarray*}
$$
but I don't see a general pattern here.
Is there a general analytical expression?
Note that this is not a textbook exercise. I'm not sure if a solution exists.
combinatorics summation binomial-coefficients
asked Jan 17 at 16:07
beckobecko
2,43131943
$endgroup$
$begingroup$
There is a formula which uses the Stirling numbers of both kinds, or Stirling numbers of the second kind and falling factorials. Use the formula $n^p = sum {pbrace k}(n)_k$, then use the technique in drhab’s answer. If you like, you can use $(N)_k=sum {kbrace i}N^i$ to get a formula in powers of $N$.
$endgroup$
– Mike Earnest
Jan 17 at 17:00
add a comment |
$begingroup$
I'm trying to find a general closed-form expression for the following sum:
$$m_p(N) = sum_{n=0}^{N} binom{N}{n} n^p$$
where $N,pge 0$ are integers.
So far I've been able to evaluate:
$$
begin{eqnarray*}
sum_n binom{N}{n} n^0 & = & 2^N\
sum_n binom{N}{n} n^1 & = & 2^{N - 1} N\
sum_n binom{N}{n} n^2 & = & 2^{N - 2} (N + 1) N\
sum_n binom{N}{n} n^3 & = & 2^{N - 3} (N + 3) N^2
end{eqnarray*}
$$
but I don't see a general pattern here.
Is there a general analytical expression?
Note that this is not a textbook exercise. I'm not sure if a solution exists.
combinatorics summation binomial-coefficients
asked Jan 17 at 16:07
beckobecko
2,43131943
$endgroup$
I'm trying to find a general closed-form expression for the following sum:
$$m_p(N) = sum_{n=0}^{N} binom{N}{n} n^p$$
where $N,pge 0$ are integers.
So far I've been able to evaluate:
$$
begin{eqnarray*}
sum_n binom{N}{n} n^0 & = & 2^N\
sum_n binom{N}{n} n^1 & = & 2^{N - 1} N\
sum_n binom{N}{n} n^2 & = & 2^{N - 2} (N + 1) N\
sum_n binom{N}{n} n^3 & = & 2^{N - 3} (N + 3) N^2
end{eqnarray*}
$$
but I don't see a general pattern here.
Is there a general analytical expression?
Note that this is not a textbook exercise. I'm not sure if a solution exists.
combinatorics summation binomial-coefficients
combinatorics summation binomial-coefficients
asked Jan 17 at 16:07
beckobecko
2,43131943
asked Jan 17 at 16:07
beckobecko
2,43131943
asked Jan 17 at 16:07
beckobecko
2,43131943
asked Jan 17 at 16:07
beckobecko
2,43131943
asked Jan 17 at 16:07
beckobecko
2,43131943
2,43131943
$begingroup$
There is a formula which uses the Stirling numbers of both kinds, or Stirling numbers of the second kind and falling factorials. Use the formula $n^p = sum {pbrace k}(n)_k$, then use the technique in drhab’s answer. If you like, you can use $(N)_k=sum {kbrace i}N^i$ to get a formula in powers of $N$.
$endgroup$
– Mike Earnest
Jan 17 at 17:00
add a comment |
$begingroup$
There is a formula which uses the Stirling numbers of both kinds, or Stirling numbers of the second kind and falling factorials. Use the formula $n^p = sum {pbrace k}(n)_k$, then use the technique in drhab’s answer. If you like, you can use $(N)_k=sum {kbrace i}N^i$ to get a formula in powers of $N$.
$endgroup$
– Mike Earnest
Jan 17 at 17:00
$begingroup$
There is a formula which uses the Stirling numbers of both kinds, or Stirling numbers of the second kind and falling factorials. Use the formula $n^p = sum {pbrace k}(n)_k$, then use the technique in drhab’s answer. If you like, you can use $(N)_k=sum {kbrace i}N^i$ to get a formula in powers of $N$.
$endgroup$
– Mike Earnest
Jan 17 at 17:00
$begingroup$
There is a formula which uses the Stirling numbers of both kinds, or Stirling numbers of the second kind and falling factorials. Use the formula $n^p = sum {pbrace k}(n)_k$, then use the technique in drhab’s answer. If you like, you can use $(N)_k=sum {kbrace i}N^i$ to get a formula in powers of $N$.
$endgroup$
– Mike Earnest
Jan 17 at 17:00
add a comment |
1 Answer
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The following equality might be helpful:
$$sum_nbinom{N}{n}left(nright)_{p}=2^{N-p}left(Nright)_{p}$$
Here $left(xright)_{p}$ is a notation for $xleft(x-1right)cdotsleft(x-p+1right)$.
answered Jan 17 at 16:25
drhabdrhab
104k545136
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add a comment |
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1 Answer
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active
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1 Answer
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active
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$begingroup$
The following equality might be helpful:
$$sum_nbinom{N}{n}left(nright)_{p}=2^{N-p}left(Nright)_{p}$$
Here $left(xright)_{p}$ is a notation for $xleft(x-1right)cdotsleft(x-p+1right)$.
answered Jan 17 at 16:25
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
The following equality might be helpful:
$$sum_nbinom{N}{n}left(nright)_{p}=2^{N-p}left(Nright)_{p}$$
Here $left(xright)_{p}$ is a notation for $xleft(x-1right)cdotsleft(x-p+1right)$.
answered Jan 17 at 16:25
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
The following equality might be helpful:
$$sum_nbinom{N}{n}left(nright)_{p}=2^{N-p}left(Nright)_{p}$$
Here $left(xright)_{p}$ is a notation for $xleft(x-1right)cdotsleft(x-p+1right)$.
answered Jan 17 at 16:25
drhabdrhab
104k545136
$endgroup$
The following equality might be helpful:
$$sum_nbinom{N}{n}left(nright)_{p}=2^{N-p}left(Nright)_{p}$$
Here $left(xright)_{p}$ is a notation for $xleft(x-1right)cdotsleft(x-p+1right)$.
answered Jan 17 at 16:25
drhabdrhab
104k545136
answered Jan 17 at 16:25
drhabdrhab
104k545136
answered Jan 17 at 16:25
drhabdrhab
104k545136
answered Jan 17 at 16:25
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
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There is a formula which uses the Stirling numbers of both kinds, or Stirling numbers of the second kind and falling factorials. Use the formula $n^p = sum {pbrace k}(n)_k$, then use the technique in drhab’s answer. If you like, you can use $(N)_k=sum {kbrace i}N^i$ to get a formula in powers of $N$.
$endgroup$
– Mike Earnest
Jan 17 at 17:00