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I've got an expression which is sum of products like:

$$a_1 a_2 + b_1 b_2 + c_1 c_2 + \cdots,$$

but the real problem I'm solving could be easily solved if I could convert this expression into something like :

$$(a_1+b_1+c_1+\cdots) \cdot (a_2+b_2+c_2+\cdots)$$

If some additional constants are added or subtracted, it's no problem, in fact, it's obvious.

First query: Can we convert it into something like required form?

Second query: If yes, how?

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    $\begingroup$ In general there are so identities of this form (for a reasonably interpretation of "this form"). Of course, the first expression is just the formula for the usual dot product on, e.g., $\Bbb R^n$ of vectors $(a_1, b_1, \ldots)$ and $(a_2, b_2, \ldots)$. $\endgroup$ Sep 13, 2015 at 4:13
  • $\begingroup$ Then, how do we get second expression, I mean, is there some standard identity, to get it? $\endgroup$ Sep 13, 2015 at 5:16
  • $\begingroup$ There is no way to get the second expression from the first without just adding in the missing terms (the so-called 'cross terms'). $\endgroup$ Sep 13, 2015 at 5:18
  • $\begingroup$ Oh, I get it, you are saying that it's easy to get first expression from second one, but not second from the first one because of the missing terms. Like for the simplest example: (a+b)*(a+b) can result in aa+bb+2*ab, but aa+b*b can not result into anything reasonable, right? $\endgroup$ Sep 13, 2015 at 5:23
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    $\begingroup$ Not so much the first from the second, but one can always write a product of sums as the sum of products just by distributing. But yes, in general we cannot write, e.g., $a_1 a_2 + b_1 b_2$ in some simpler way. $\endgroup$ Sep 13, 2015 at 5:28

1 Answer 1

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In general, one can write a product of sums as a sum of a products: $$\left(\sum_{i \in I} x_i\right)\left(\sum_{j \in J} y_j \right) = \sum_{i \in I, j \in J} x_i y_j.$$ One cannot, however, in general reverse this process, that is, write a sum of products, $$\phantom{(\ast) \qquad} \sum_{k \in K} x_k y_k, \qquad (\ast)$$ as a product of sums. (In this answer we assume that the index sets, $I, J$, etc., are finite, though with some care we can extend them to infinite sets under suitable conditions.) Note that the factors $x_k, y_k$ of each summand in $(\ast)$ are indexed by the same set $K$, whereas that is not (generally) the case for the sum of products in the first display equation. When they are, we can write the sum of products in terms of a product of sums with a correction term, namely as $$\sum_{k \in K} x_k y_k = \left(\sum_{k \in K} x_k\right) \left(\sum_{k' \in K} y_{k'}\right) - \sum_{k, k' \in K; k \neq k'} x_k y_{k'},$$ but this is really just a reorganization, and not really an algebraic simplification.

The expression $(\ast)$ can be factored in the sense that it is precisely the standard "dot product" on the space $\oplus_{k \in K} R$ of ordered $|K|$-tuples (with components indexed by some finite set $K$) of ${\bf x} := (x_k)$ and ${\bf y} := (y_k)$ with entries in some ring $R$, $${\bf x} \cdot {\bf y} := \sum_{k \in K} x_k y_k,$$ though as the notation suggests, this is a definition and again not an simplification per se.

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  • $\begingroup$ Thank you. Where I can find more about this? Can you recommend a book for me please? $\endgroup$
    – Avv
    Mar 19, 2021 at 0:58
  • $\begingroup$ I'm not sure what to recommend exactly, since as written this post is pretty narrow. Can you provide some more context about what you're looking for and why? $\endgroup$ Mar 19, 2021 at 1:14
  • $\begingroup$ I found similar concepts and questions in my Calculus book. Thank you. $\endgroup$
    – Avv
    Mar 19, 2021 at 14:41
  • $\begingroup$ it seems like the distinction between the two directions is that when you sum over $i,j \in K$, you end up with extra copies of the terms $x_k y_{k'}$ and $x_{k'} y_k$ for $k\neq k'$? Why is this the case? For what sets $K$ does the backward implication not hold, and for what $K$ does it hold? $\endgroup$
    – 900edges
    Mar 5 at 19:28
  • $\begingroup$ @ladygaga That's just the difference between using a sum over a single index and a double sum over $2$ indices. In general, $\sum_{k \in K} \sum_{k' \in K} x_k y_{k'}$ coincides with $\sum_{k \in K} x_k y_k$ (regarded as abstract polynomials) if and only if the "correction term" in the third display equation in my answer is zero, that is, if $|K| \leq 1$. $\endgroup$ Mar 5 at 20:03

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