I was wondering is it possible to simplify the following sum:
$$\sum_{i_1+i_2+...+i_m=z}{x_{1}^{i_1}x_{2}^{i_2}\cdot...\cdot x_{m}^{i_m}}$$
where $0<x<1$ for all $x$.
Is it possible to lose the sum? For $m=2$ it is simple. Just could not find it for $m>2$. Maybe it has to do with multinomial theorem, but how?
Would appreciate your help.
$$m, z, i_1, \ldots, i_m \in \mathbb{N}, \quad m > 0$$
I don't think your sum has a neat form. Let $m = z = 2$ your "simple form" is: $x_1^2 + x_1\cdot x_2 + x_2^2$, which is not the binomial formula. Your formula only counts the mixed terms once. The multinomial formula has the following form:
$$\left(\sum\limits_{k=1}^m x_i\right)^z = \sum\limits_{\sum\limits_{k=1}^m i_k=z}\frac{z!}{\prod\limits_{l=1}^m i_l!}\prod\limits_{l=1}^m x_l^{i_l} = \sum\limits_{i_1 + \ldots + i_m = z}\frac{z!}{i_1!\cdot \ldots \cdot i_m!}x_1^{i_1}\cdot\ldots\cdot x_m^{i_m}$$