Notation of sigmas

Question

When one writes $$\sum_{i+j=4} a_i a_j$$ is that then equal to $$a_0a_4 + a_1a_3 + a_2^2$$ or $$\sum_{i=0}^4 a_i a_{4-i}=2a_0a_4+2a_1a_3 + a_2^2$$ I suddenly got confused while writing these sigmas down. Perhaps it depends on context or perhaps not? Thanks for any help.

Answer 1

I think that there is no ambiguity on the sum. When you write

$$ \sum_{i+j = 4} a_i a_j$$

you are summing over the set of all solutions of $i+j = 4$. The solutions are

• $i = 0$ and $j = 4$
• $i = 1$ and $j = 3$
• $i = 2$ and $j = 2$
• $i = 3$ and $j = 1$
• $i = 4$ and $j = 0$

So the sum should result:

$$ \sum_{i+j = 4} a_i a_j = a_0a_4 + a_1a_3 + a_2a_2 + a_3a_1 + a_4a_0 = 2a_0a_4 + 2a_1a_3 + a_2^2$$

The confusions only happens because you are indexing $a$ twice, how would you interpret the following sum? $$ \sum_{i+j = 4} a_i b_j$$

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Most of the places where sums like this appears on the indexing set of the summation, it on a product of generating functions / power series, like this:

$$ \left ( \sum_{i=0}^\infty a_i x^i \right ) \left ( \sum_{j=0}^\infty b_j x^j \right ) = \sum_{n=0}^\infty \left ( \sum_{i+j = n} a_i b_j \right )x^n $$

Note that the coefficient of $x^n$ on the right-hand side is a sum like yours. The correct interpretation on this case should be:

$$ \sum_{i+j = n} a_i b_j = a_0 b_n + a_1 b_{n-1} + a_2 b_{n-2} + \dotsc + a_n b_0 $$

This also expands nicely for multiple variables, so we have things like the multinomial theorem

$$\left ( x_1 + x_2 + \dotsc + x_m \right )^n = \sum_{k_1 + k_2 + \dotsc + k_m = n} \binom{n}{k_1,k_2,\dotsc,k_m} \prod_{t=1}^{m}x_t^{k_t}$$

here the sum have to be interpreted as being over all the $m$-tuples $(k_1, k_2, \dotsc, k_m)$ satisfying the equation. So this is the standard interpretation for sums of this type.

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Of course that maybe you want to sum over the set that you have presented, but then I think is best to indicate explicitly, as that's not the common case. One way to do it would be:

$$\sum_{\substack{i+j = 4 \\ i \leq j}} a_i a_j = a_0a_4 + a_1 a_3 + a_2^2$$

Note that this can also be easily generalized.