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.

  • $\begingroup$ What do you mean "for $m=2$ it is simple"? Writing as quotient per geometric series formula? $\endgroup$ – Hagen von Eitzen Jan 8 at 23:33
  • $\begingroup$ if $z=m$ then I think it is the expansion of $(x_1+x_2+x_3+\cdots+x_m)^m$ $\endgroup$ – Shrey Joshi Jan 8 at 23:53
  • $\begingroup$ @Shrey: no its not, because in the case $m = z = 2$ it counts mixed terms like $x_1x_2$ only once. $\endgroup$ – Snake707 Jan 9 at 0:00
  • 1
    $\begingroup$ Your expression is known as the complete homogeneous symmetric polynomial of degree z in m variables. $\endgroup$ – Mike Earnest Jan 9 at 0:31
  • $\begingroup$ Thank you so much @MikeEarnest! you've just solved my problem. Thank god for crowd wisdom :) $\endgroup$ – Y.L Jan 9 at 2:13

$$m, z, i_1, \ldots, i_m \in \mathbb{N}, \quad m > 0$$

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:

$$\left(\sum\limits_{k=1}^m x_i\right)^z = \sum\limits_{\sum\limits_{k=1}^m i_k=z}\frac{z!}{\prod\limits_{l=1}^m i_l!}\prod\limits_{l=1}^m x_l^{i_l} = \sum\limits_{i_1 + \ldots + i_m = z}\frac{z!}{i_1!\cdot \ldots \cdot i_m!}x_1^{i_1}\cdot\ldots\cdot x_m^{i_m}$$


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