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The following is from Knuth's "Concrete Mathematics",

$$\begin{align}\sum_{1 \le j, k \le 3} a_j b_k = a_1 b_1 + &a_1 b_2 + a_1 b_3 \\ + a_2 &b_1 + a_2 b_2 + a_2 b_3 \\ + a_3 &b_1 + a_3 b_2 + a_3 b_3\end{align}$$

Would it be possible to use the distributive law:

$a_j * \sum_{k \leq 3} b_k$

Or is it only possible to remove a constant $c$ in a distribution such as if we had $\sum_{k \leq 3} c*b_k$

Then it would be $c * \sum_{k \leq 3} b_k$?

Katie Melosto
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1 Answers1

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You could rewrite it as $$\sum\limits_{\substack{1\le j \\ k\le3}} a_jb_k = \left(\sum_{1\le j}a_j\right)\cdot\left(\sum_{k\le 3} b_k\right)$$

Which, of course, is a sum multiplied by a sum, so the distribution takes place, resulting in the right hand side given by Knuth.

MonsterEater
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