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I'm studying tensors right now and there are lots of sums involved and I've seen they do these sorts of things:

$(\sum_i x)(\sum_j y) (\sum_k z)=\sum_k \sum_j \sum_i x y z$

I've shown mechanically that this is true due to the distributive property when I consider two split sums. But I want a general proof that this way of joining the sums is valid and makes sense. I guess I'm not specifically looking for an equivalence to the distributive property, but just a statement that justifies such a step.

Thanks a whole lot.

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  • $\begingroup$ Note that this only follows from the distributive law if the sums are finite. For infinite sums or integrals you need to make some extra "niceness" assumptions about the sequences or functions being summed. $\endgroup$ – Mario Carneiro Feb 21 '15 at 3:27
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You can "see" it by noticing that both sides are polynomials in the $x_i,y_j,z_k$ and put all of them equal zero except for one particular triple $x_{i_0},y_{j_0},z_{k_0}$. Both sides become $x_{i_0}y_{j_0}z_{k_0}$.

Don't take this as the proof. Just a a way to get comfortable about the identity.

To prove it use distributibity.

$$\begin{align}(\sum_i x_i)(\sum_j x_j)(\sum_k z_k)&=\sum_k\left[ (\sum_i x_i)(\sum_j y_j)z_k\right]\\&=\sum_k\left[ \sum_j\left\{(\sum_i x_i)y_jz_k\right\}\right]\\&=\sum_k\left[ \sum_j\left\{\sum_i x_iy_jz_k\right\}\right]\end{align}$$

Associativity and commutativity allows you to put the parentheses and terms in any order. The step-by-step deduction from the $(a+b)+c=a+(b+c)$ definition I refuse to do.

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Yes, this is just the distributivity of multiplication over addition. Once you prove it for two summations it extends to any finite number by repeated applications of $$(\sum_i x)(\sum_j y) (\sum_k z)=(\sum_i x)\sum_k \sum_j y z=\sum_k \sum_j \sum_i x y z$$ You just deal with one variable at a time.

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