Analytical form of the binomial moments












0












$begingroup$


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.










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  • $begingroup$
    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
















0












$begingroup$


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.










share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0


1



$begingroup$


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.










share|cite|improve this question









$endgroup$




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






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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


















  • $begingroup$
    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










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 $xleft(x-1right)cdotsleft(x-p+1right)$.






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    1 Answer
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    1 Answer
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    active

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    0












    $begingroup$

    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 $xleft(x-1right)cdotsleft(x-p+1right)$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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 $xleft(x-1right)cdotsleft(x-p+1right)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $xleft(x-1right)cdotsleft(x-p+1right)$.






        share|cite|improve this answer









        $endgroup$



        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 $xleft(x-1right)cdotsleft(x-p+1right)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 17 at 16:25









        drhabdrhab

        104k545136




        104k545136






























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