0
$\begingroup$

I encountered two Laurent polynomials $$\sum_{j=0}^k\sum_{ \mbox{$\begin{array}{c} i_1,\dots,i_k\geq 0\\ i_0=-N-i_1-\dots-i_k\end{array}$}} \sum_{m=0}^{i_j} \begin{pmatrix} -N\\ i_1,\dots,i_k \end{pmatrix} \begin{pmatrix} i_j\\ m \end{pmatrix} x_0^{i_0}\cdots x^{i_j-m}\cdots x_k^{i_k} $$ and
$$ \sum_{j=0}^k\sum_{N\geq p_1\geq\cdots\geq p_{n}\geq 0} \sum_{ \mbox{$\begin{array}{c} i_1,\dots,i_k\geq 0\\ i_0=-N-i_1-\dots-i_k-p_1-\cdots- p_{n}\end{array}$}} \begin{pmatrix} N\\ p_1 \end{pmatrix} \begin{pmatrix} p_1\\ p_2 \end{pmatrix}\cdots \begin{pmatrix} p_{n-1}\\ p_{n} \end{pmatrix} \begin{pmatrix} -(N+p_1+\cdots+p_n)\\ i_1,\dots,i_k \end{pmatrix} x_0^{i_0}\cdots x^{i_j+p_1+\cdots+p_n}\cdots x_k^{i_k}, $$ where $N,n,k$ are given positive integers and $$ \begin{pmatrix} l\\ i_1,\dots,i_k \end{pmatrix}= \begin{pmatrix} l\\ i_1 \end{pmatrix} \begin{pmatrix} l-i_1\\ i_2 \end{pmatrix}\cdots \begin{pmatrix} l-i_1-\cdots-i_{k-1}\\ i_k \end{pmatrix} $$ for any integer $l$, positive integers $i_1,\dots,i_k$.
I need to prove that they are actually the same.
Can anyone help me? or just give me some hint. I'll be grateful.

$\endgroup$
2
  • $\begingroup$ Did you check that they are the same? (Say, by a computer experiment.) Accidentally, what are the parameters? $N$ and $k$? At first sight, your "polynomials" look more like power series. $\endgroup$ Mar 14, 2014 at 12:30
  • $\begingroup$ There is no $i_0$, so $j$ should run in both sums from 1 to k. Using $\sum_{m=0}^{i} \binom imx^{i-m}=\sum_{m=0}^{i} \binom imx^{m}=(1+x)^i$ with $i=i_j$ repeatedly for each $j$, you should come up with some power of $(k+x)$ for both expressions. I wonder though if your summations are well-defined. $\endgroup$
    – Wolfgang
    Mar 14, 2014 at 18:09

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.