# A question concerning multi-indices

I am having difficulties understanding the following formula : $$(x_1+\cdots+x_n)^k=\sum_{\alpha,|\alpha|=k}\frac{|\alpha|!}{\alpha!}x^\alpha$$ where $\alpha$ is a multi-index.

I find this notation very confusing, I can't even evaluate the first term of the sum.

My question is why does this formula hold and how can the sum be expanded?

Here is what I know :

$\alpha =(\alpha_1,\alpha_2,\cdots,\alpha_n)$ , $|\alpha|=\alpha_1+\alpha_2+\cdots+\alpha_n$ and $\alpha!=\alpha_1!\alpha_2!\cdots\alpha_n!$

Clearly expanding the sum with these yeilds the multinomial expression. What I cant seem to understand is how does the sum expand.

Edit: I think it makes sense now.

What I needed to know ( and well didn't :/): The condition $|\alpha|=k$ means $\alpha$ is fixed and all that needs to to be done is to find some $\alpha_i's$ whose sum is $k$ and the number of $\alpha_i's$ is equal to the number of terms ($n$ in the equation above)

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For $k$ a non-negative integer write it as sum of $n$ non-negative numbers. The numerator is just $k!$ and denominator is the product of factorial of the terms of your sum. –  Maesumi Mar 3 '13 at 15:59

It means $$(x_1 + x_2 + \cdots + x_n)^k = \sum_{k_1+k_2+\cdots+k_n=k} {k \choose k_1, k_2, \ldots, k_n} x_1^{k_1} x_2^{k_2} \cdots x_n^{k_n},$$ where $${k \choose k_1, k_2, \ldots, k_n} = \frac{k!}{k_1!\, k_2! \cdots k_n!}$$ are the multinomial coefficients. In the same Wikipedia article you can find a proof, which generalizes that of the binomial theorem.
For instance $$(x_1 + x_2 + x_3)^3$$ can be expanded as follows, where the terms correspond to the values of $(k_1,k_2,k_3)$ given by $(3,0,0), (2,1,0), (2, 0, 1), (1,2,0), (1,1,1), (1,0,2),(0,3,0), (0,2,1), (0,1,2),(0,0,3)$ $$x_1^3 + 3 x_1^2 x_2 + 3 x_1^2 x_3 + 3 x_1 x_2^2 + 6 x_1 x_2 x_3 + 3 x_1 x_3^2 + x_2^3 + 3 x_2^2 x_3 + 3 x_2 x_3^2 + x_3^3 .$$ Here $$\frac{3!}{3!\, 0!\, 0!} = 1, \quad \frac{3!}{2!\, 1!\, 0!} = 3, \quad \frac{3!}{1!\, 1!\, 1!} = 6.$$
Ok so you used $\alpha =(k_1,k_2, \cdots, k_n)$ and expanded, but I still do not understand how the multinomial coefficients are determined. –  user10444 Mar 3 '13 at 16:05
@user10444, I have seen your reformulated question. Please try to work out manually some small cases, first in the binomial case $n=2$, and then $n=3$, and you'll see how the binomial/multinomial coefficients pop up. –  Andreas Caranti Mar 3 '13 at 16:13
I'll try $(x_1+x_2+x_3)^3$ but is there no way to calculate them in general? –  user10444 Mar 3 '13 at 16:17