Expansion of a Polynomial

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Let $mathbf s in mathbb N^N$ be a fixed vector, and define the polynomial $a(cdot)$ such that
beginequation
a(mathbf x)=prod_i=1^Nbinomx_is_i
endequation
for all $mathbf x in mathbb N^N$, where $binomx_is_i = fracx_i!s_i!(x_i - s_i)!$.



I'd like to write this polynomial as a linear combination of monomials. In particular, I'd like to write
beginequation
a(mathbf x) = sum_mathbf j in mathbb N^N b_mathbf j mathbf x^mathbf j,
endequation
where $mathbf x^mathbf j = prod_i = 1^N x_i^j_i$, and I have analytical expressions for the each coefficient $b_mathbf j$ in terms of the vector $mathbf s$.



Does anybody know how to do this?




combinatorics polynomials




share|cite|improve this question

















  • 1




    According to Wikipedia's article on binomial coefficients, $binomx_is_i$ can be expanded in terms of "Stirling numbers of the first kind".
    – Garrett
    Jul 27 at 20:22











  • en.wikipedia.org/wiki/Binomial_coefficient
    – Garrett
    Jul 27 at 20:37










  • That's probably the key fact I needed.
    – Garrett
    Jul 27 at 20:38














up vote
1
down vote

favorite
1












Let $mathbf s in mathbb N^N$ be a fixed vector, and define the polynomial $a(cdot)$ such that
beginequation
a(mathbf x)=prod_i=1^Nbinomx_is_i
endequation
for all $mathbf x in mathbb N^N$, where $binomx_is_i = fracx_i!s_i!(x_i - s_i)!$.



I'd like to write this polynomial as a linear combination of monomials. In particular, I'd like to write
beginequation
a(mathbf x) = sum_mathbf j in mathbb N^N b_mathbf j mathbf x^mathbf j,
endequation
where $mathbf x^mathbf j = prod_i = 1^N x_i^j_i$, and I have analytical expressions for the each coefficient $b_mathbf j$ in terms of the vector $mathbf s$.



Does anybody know how to do this?




combinatorics polynomials




share|cite|improve this question

















  • 1




    According to Wikipedia's article on binomial coefficients, $binomx_is_i$ can be expanded in terms of "Stirling numbers of the first kind".
    – Garrett
    Jul 27 at 20:22











  • en.wikipedia.org/wiki/Binomial_coefficient
    – Garrett
    Jul 27 at 20:37










  • That's probably the key fact I needed.
    – Garrett
    Jul 27 at 20:38












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Let $mathbf s in mathbb N^N$ be a fixed vector, and define the polynomial $a(cdot)$ such that
beginequation
a(mathbf x)=prod_i=1^Nbinomx_is_i
endequation
for all $mathbf x in mathbb N^N$, where $binomx_is_i = fracx_i!s_i!(x_i - s_i)!$.



I'd like to write this polynomial as a linear combination of monomials. In particular, I'd like to write
beginequation
a(mathbf x) = sum_mathbf j in mathbb N^N b_mathbf j mathbf x^mathbf j,
endequation
where $mathbf x^mathbf j = prod_i = 1^N x_i^j_i$, and I have analytical expressions for the each coefficient $b_mathbf j$ in terms of the vector $mathbf s$.



Does anybody know how to do this?




combinatorics polynomials




share|cite|improve this question













Let $mathbf s in mathbb N^N$ be a fixed vector, and define the polynomial $a(cdot)$ such that
beginequation
a(mathbf x)=prod_i=1^Nbinomx_is_i
endequation
for all $mathbf x in mathbb N^N$, where $binomx_is_i = fracx_i!s_i!(x_i - s_i)!$.



I'd like to write this polynomial as a linear combination of monomials. In particular, I'd like to write
beginequation
a(mathbf x) = sum_mathbf j in mathbb N^N b_mathbf j mathbf x^mathbf j,
endequation
where $mathbf x^mathbf j = prod_i = 1^N x_i^j_i$, and I have analytical expressions for the each coefficient $b_mathbf j$ in terms of the vector $mathbf s$.



Does anybody know how to do this?





combinatorics polynomials





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 27 at 19:25
























asked Jul 27 at 19:02









Garrett

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536314







  • 1




    According to Wikipedia's article on binomial coefficients, $binomx_is_i$ can be expanded in terms of "Stirling numbers of the first kind".
    – Garrett
    Jul 27 at 20:22











  • en.wikipedia.org/wiki/Binomial_coefficient
    – Garrett
    Jul 27 at 20:37










  • That's probably the key fact I needed.
    – Garrett
    Jul 27 at 20:38












  • 1




    According to Wikipedia's article on binomial coefficients, $binomx_is_i$ can be expanded in terms of "Stirling numbers of the first kind".
    – Garrett
    Jul 27 at 20:22











  • en.wikipedia.org/wiki/Binomial_coefficient
    – Garrett
    Jul 27 at 20:37










  • That's probably the key fact I needed.
    – Garrett
    Jul 27 at 20:38







1




1




According to Wikipedia's article on binomial coefficients, $binomx_is_i$ can be expanded in terms of "Stirling numbers of the first kind".
– Garrett
Jul 27 at 20:22





According to Wikipedia's article on binomial coefficients, $binomx_is_i$ can be expanded in terms of "Stirling numbers of the first kind".
– Garrett
Jul 27 at 20:22













en.wikipedia.org/wiki/Binomial_coefficient
– Garrett
Jul 27 at 20:37




en.wikipedia.org/wiki/Binomial_coefficient
– Garrett
Jul 27 at 20:37












That's probably the key fact I needed.
– Garrett
Jul 27 at 20:38




That's probably the key fact I needed.
– Garrett
Jul 27 at 20:38















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