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I would like simplify the following doble sum

$$ \sum_{k=m}^n\,s(n,k)\,x^k\sum_{s=m}^k\,(-1)^{k+s}\,s(k,s)\begin{pmatrix}s\\m\end{pmatrix}\,y^{s-m}$$

with $s(n,k)$ the Stirling numbers of first kind. I've be able to invert the order of summation

$$ \sum_{s=m}^n\begin{pmatrix}s\\m\end{pmatrix}\,y^{s-m}\sum_{k=s}^n\,(-1)^{k+s}\,s(n,k)\,s(k,s)\,x^k$$

but the difficulty seems to be the same.

Any help will be welcomed.

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  • $\begingroup$ Usually, the lower case $s(n,k)=(-1)^{n-k}\left[n\atop k\right]$ is used for the signed Stirling numbers of the first kind; is that what you intended? $\endgroup$ – robjohn Apr 2 at 23:15
  • $\begingroup$ Yes, you are right. I thought this notation was usual. $\endgroup$ – popi Apr 2 at 23:38
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Let me change a bit the symbols and write $$ \eqalign{ & M_{\,n,\,m} (x,y) = \sum\limits_{k = m}^n { \left[ \matrix{ n \cr k \cr} \right]x^{\,k} \sum\limits_{j = m}^k { \left( { - 1} \right)^{\,k + j} \left[ \matrix{ k \cr j \cr} \right]\left( \matrix{ j \cr m \cr} \right)y^{\,j - m} } } = \cr & = \sum\limits_{\left( {m\, \le } \right)\,k\,\left( { \le \,n} \right)} { \sum\limits_{\left( {m\, \le } \right)\,j\,\left( { \le \,k} \right)} { \left( { - 1} \right)^{\,k + j} \left[ \matrix{ n \cr k \cr} \right]\left[ \matrix{ k \cr j \cr} \right] \left( \matrix{ j \cr m \cr} \right)x^{\,k} y^{\,j - m} } } = \cr & = \sum\limits_{\left( {m\, \le } \right)\,k\,\left( { \le \,n} \right)} { \sum\limits_{\left( {m\, \le } \right)\,j\,\left( { \le \,k} \right)} { \left[ \matrix{ n \cr k \cr} \right]x^{\,k} \left( { - 1} \right)^{\,k} \left[ \matrix{ k \cr j \cr} \right] \left( { - 1} \right)^{\,j} y^{\,j} \left( \matrix{ j \cr m \cr} \right)y^{\, - m} } } \tag{1} \cr} $$ where the bounds of the sums are put in brackets to show that they are actually implicit in the binomial and Stirling 1st kind.

Now we have that $$ \eqalign{ & x^{\,\overline {\,n\,} } = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} { \left[ \matrix{ n \cr k \cr} \right]x^{\,k} } \quad \Rightarrow \cr & \Rightarrow \quad \left( {x + y} \right)^{\,\overline {\,k\,} } = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} { \left[ \matrix{ k \cr j \cr} \right]\left( {x + y} \right)^{\,j} } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} { \sum\limits_{\left( {0\, \le } \right)\,m\,\left( { \le \,j} \right)} { \left[ \matrix{ k \cr j \cr} \right]\left( \matrix{ j \cr m \cr} \right)x^{\,m} y^{\,j - m} } } \quad \Rightarrow \cr & \Rightarrow \quad \left( { - x - y} \right)^{\,\overline {\,k\,} } = \left( { - 1} \right)^{\,k} \left( {x + y} \right)^{\,\underline {\,k\,} } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,m\,\left( { \le \,k} \right)} { \left( {\sum\limits_{\left( {m\, \le } \right)\,j\,\left( { \le \,k} \right)} { \left[ \matrix{ k \cr j \cr} \right]\left( \matrix{ j \cr m \cr} \right)\left( { - 1} \right)^{\,j} x^{\,m} y^{\,j - m} } } \right)} \quad \Rightarrow \cr & \Rightarrow \quad \sum\limits_{\left( {m\, \le } \right)\,j\,\left( { \le \,k} \right)} { \left[ \matrix{ k \cr j \cr} \right]\left( \matrix{ j \cr m \cr} \right)\left( { - 1} \right)^{\,j} y^{\,j - m} } = \left[ {x^{\,m} } \right]\left( { - x - y} \right)^{\,\overline {\,k\,} } \cr} $$ Therefore $$ \eqalign{ & M_{\,n,\,m} (x,y) = \cr & = \sum\limits_{\left( {m\, \le } \right)\,k\,\left( { \le \,n} \right)} { \left[ \matrix{ n \cr k \cr} \right]x^{\,k} \left( {\left[ {x^{\,m} } \right]\left( {x + y} \right)^{\,\underline {\,k\,} } } \right)} = \cr & = {1 \over {m!}}\sum\limits_{\left( {m\, \le } \right)\,k\,\left( { \le \,n} \right)} { \left[ \matrix{ n \cr k \cr} \right]x^{\,k} {{d^{\,m} } \over {dy^{\,m} }}y^{\,\underline {\,k\,} } } \cr \tag{2}} $$

where $\left[ {x^{\,m} } \right]$ stands for "the coefficient of $x^m$ (in the following expression)".

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  • $\begingroup$ Many thanks!! Only a minor question, what is the meanig of $[x^m]$ in the two Last pharagraphs? $\endgroup$ – popi Apr 2 at 21:42
  • $\begingroup$ sorry, I added the explanation for the symbol (see for instance this) $\endgroup$ – G Cab Apr 2 at 22:44
  • $\begingroup$ Thanks for the appreciation, hope the answer helped you $\endgroup$ – G Cab Apr 7 at 21:42

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