# 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:

$$\sum_{|\alpha| \le\ 2}a_\alpha x^\alpha, x \in\ \mathbb{R}^2$$ Which I understand expands to:
$$= \sum_{|\alpha| =\ 0}a_\alpha x^\alpha + \sum_{|\alpha| = 1}a_\alpha x^\alpha + \sum_{|\alpha| = 2}a_\alpha x^\alpha$$ And here where I am lost : $$= a_{00}x_1^0x_2^0 + a_{10}x_1^1x_2^0 + a_{01}x_1^0x_2^1 + a_{11}x_1^1x_2^1 + a_{20}x_1^2x_2^0 + a_{02}x_1^0x_2^2$$

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.

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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: $(i,j) \in \{(0,2), (1,1), (2,0)\}$.

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Thanks for taking the time. I get it now. – emilyjac Jun 12 '13 at 6:51
@emilyjac no problem, happy to help you. – gt6989b Jun 12 '13 at 12:39