2
$\begingroup$

I'm trying to simplify the sum $$ \sum_{\vec x \in (\mathbb{N}_0)^n: M\vec x = \vec b} \prod_i \frac{(a_i)^{x_i}}{x_i!}, $$ where $M$ is a $\mathbb{N}_0$-valued $m\times n$ matrix, $\vec b$ is $\mathbb{N}_0$-valued length-$m$ vector, and $a_i$ are real numbers. In my specific use case (if that makes it simpler by any chance), the $a_i$ are 4th roots of unity, i.e., $1$, $i$, $-1$, or $-i$, and the set of $\vec x$ fulfilling $M\vec x=\vec b$ is finite.

I found using Mathematica that for a special case of this, we have the very simple formula $$ \sum_{x_1, x_2: x_1+x_2=y} \frac{a^{x_1} b^{x_2}}{x_1!x_2!} = \frac{(a+b)^y}{y!}. $$ This way we can get rid of one variable $x_i$ which is involved in only one constraint in $M$. I thought maybe one could do some sort of Gaussian elimination on $M$ to use this for solving the general case, but this is only a vague idea so far.

$\endgroup$

1 Answer 1

2
$\begingroup$

Such simplification seems highly unlikely. Even in the very simply particular case when $m=2$, $n=3$, $M=\begin{pmatrix} 1&1&0 \\ 0&1&1\end{pmatrix}$, $\vec b=\begin{pmatrix} b \\ b\end{pmatrix}$, and $a_1=a_2=a_3=1$, Mathematica can only give a tautological answer:

enter image description here


Here is an even simpler case, when $m=1$, $n=2$, $M=\begin{pmatrix} 1&2\end{pmatrix}$, $\vec b=\begin{pmatrix} 2b\end{pmatrix}$, and $a_1=a_2=1$. Here too, Mathematica cannot do anything:

enter image description here

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.