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)
Best Answer
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 $$ \frac{3!}{3!\, 0!\, 0!} = 1, \quad \frac{3!}{2!\, 1!\, 0!} = 3, \quad \frac{3!}{1!\, 1!\, 1!} = 6. $$