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Is this true? $$ \sum\limits_{i<j} x_ix_j = \sum\limits_{j=1}^n \sum\limits_{i=1}^j x_ix_j$$ $i, j = 1,\ldots,n$

And on the left hand side, how can you tell when it stops?

Robert Cardona
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Benck
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  • it is usually implied that $j$ goes up to some value $n$ which should be clear from the context. It may be the dimension of a matrix or a vector for example – Ant Feb 04 '15 at 22:40
  • That's usually shorthand, with the assumption of the start and end values. You'll sometimes see the more verbose: $$\sum_{1\leq i<j\leq n} x_ix_j$$ – Thomas Andrews Feb 04 '15 at 22:41

4 Answers4

3

Not quite: you are summing terms $x_jx_j$ on the right that don't appear on the left... The correct formula is $$\sum_{j=2}^n\sum_{i=1}^{j-1}\,x_ix_j$$

Olivier Bégassat
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You are close! Notice that $\sum\limits_{j=1}^n \sum\limits_{i=1}^j x_ix_j$ contains the term $x_jx_j\forall j \leq n$

$$ \sum\limits_{i<j} x_ix_j = \sum\limits_{j=2}^n \sum\limits_{i=1}^{j-1} x_ix_j$$

jameselmore
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I like to use Iverson summation: $$ \sum_{i<j}x_ix_j=\sum_{j=1}^n\sum_{i=1}^n[i<j] x_ix_j=\sum_{j=1}^n\sum_{i=1}^{j-1}x_ix_j=\sum_{j=2}^n\sum_{i=1}^{j-1}x_ix_j $$ where $[i<j]$ is an Iverson bracket equals one if $i<j$ and zero otherwise.

user103828
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  • In the last double summation on the right, for the case $j=1$ the sum over $i$ will run from $i=1$ up to $i = j-1 = 0$--that is, the sum will have no terms. So you can just as well start at $j=2$, with the same result in the end. (I'm not saying the formula above is wrong; technically, $j=1$ works too for the reason I just gave.) – David K Feb 04 '15 at 23:16
  • Thanks... I didn't notice that... actually was wondering if the other answers had a typo. – user103828 Feb 04 '15 at 23:21
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    It's redundant to use the indicator function on an Iverson bracket. In any event, indicator functions and Iverson brackets don't return truth values, they return $1$ or $0$. – GFauxPas Feb 05 '15 at 00:26
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You wrote $\sum\limits_{i<j} x_ix_j = \sum\limits_{j=1}^n \sum\limits_{i=1}^j x_ix_j$

To change the left side so that is the same as the right, I would change it to $\sum\limits_{1\le i\le j \le n} x_ix_j $.

marty cohen
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