Notation for product expansion

Question

Please excuse me in advance if this is the wrong type of question or too simple. I need to expand the following product for a proof: $ \left( \sum_{l=1}^{i} x_{l} \right) \left( \sum_{m=1}^{j} x_{m} \right)$

Would this be an appropriate notation or is there a more elegant way to write the following:

... we expand the product in Eq.XYZ to obtain

$ ( x_1x_1 + x_1x_2 + x_1x_3 + \ldots +x_1x_j \\ +x_2x_1 + x_2x_2 + x_2x_3 + \ldots +x_2x_j \\ \vdots \\ +x_ix_1 + x_ix_2 + x_ix_3 + \ldots +x_ix_j ) $

Edit: corrected index order

Answer 1

This notation is very clear. Personally I'd do it the same way if I had to, but I would try to stick with the sums for as long as possiple, to avoid tedious notation. It is also common to use:

$$\sum_{l,m} x_lx_m$$

As long as you make clear that $1\leq l\leq i$ and $1\leq m\leq j$.

Also beware that if the $x_l$ and $x_m$ do not commute, you should reverse the $i$'s and $j$'s in your form.

Answer 2

Without loss of generality let $i>j$. Then $\left( \sum_{l=1}^{i} x_{l} \right) \left( \sum_{m=1}^{j} x_{m} \right)= \left(\sum_{l=1}^{j} x_{l} + \sum_{l=j+1}^{i} x_{l} \right)\times \sum_{m=1}^{j} x_{m}=\left(\sum_{m=1}^{j} x_{m}\right)^2+\sum_{l=j+1}^{i} \sum_{m=1}^{j} x_{l}x_{m} $

Answer 3

Without loss of generality assume that $i\ge j$. Define $a_l=x_l$ for $1\le l\le i$ and $b_l=x_l$ for $1\le l\le j$ and $b_l=0$ for $j+1\le l \le i$. Then $$\left(\sum\limits_{l=1}^i{x_l}\right)\left(\sum\limits_{l=1}^j{x_l}\right)=\left(\sum\limits_{l=1}^i{a_l}\right)\left(\sum\limits_{l=1}^i{b_l}\right)=\sum\limits_{l=1}^i{c_l},$$ where $c_l=\sum\limits_{k=1}^l{a_k b_{l-k+1}}.$