correct expansion of a sum using multiple indexes I have looked for a similar posting but haven't found anything... but then I am also a bit unsure of how to search because I've never posted a math question before. In my introductory finite element method course the prof was introducing multiple index notation and I don't understand the way that a sum has been expanded while using this notation. The example given is:
\begin{equation}
 \sum_{|\alpha| \le\ 2}a_\alpha x^\alpha, x \in\ \mathbb{R}^2
\end{equation}
Which I understand expands to:
\begin{equation}
= \sum_{|\alpha| =\ 0}a_\alpha x^\alpha + \sum_{|\alpha| = 1}a_\alpha x^\alpha + \sum_{|\alpha| = 2}a_\alpha x^\alpha
\end{equation}
And here where I am lost :
\begin{equation}
= a_{00}x_1^0x_2^0 + a_{10}x_1^1x_2^0 + a_{01}x_1^0x_2^1 + a_{11}x_1^1x_2^1 + a_{20}x_1^2x_2^0 +  a_{02}x_1^0x_2^2
\end{equation}
Which of these termes correspond to which sum?
Why does "a" have a double index?
If $|\alpha| = 2 = \alpha_1 + \alpha_2$, what does $|\alpha|=0$ mean? 0? Empty?
How do I know the values of  $\alpha_1$ and $\alpha_2$?
Please note that it is entirely possible that I have copied some of this out wrong, the prof has nearly indecipherable handwriting. If someone can point me in the right direction it would be greatly appreciated.

It sounds like the double index under $a_{ij}$ exactly correponds to the powers $x_1^i x_2^j$. The sum in question assumes $\alpha$ has 2-digits, and $|\alpha| = 1$ means the digits sum to $1$.
For simplicity write $\alpha = (i,j), a_\alpha = a_{ij}, x^\alpha = x_1^i x_2^j$ and $|\alpha| = i+j$.
So you are right, the first step expands $|\alpha| \leq 2$ into three cases of $0,1,2$. The second step expands $|\alpha| = i+j = 0$ into the only possible case $(i,j) = (0,0)$.
The second one has them sum to one: $(i,j) \in \{(0,1), (1,0)\}$.
The third one has them sum to two: $(i,j) \in \{(0,2), (1,1), (2,0)\}$.
