0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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