Can the Double Summation be Simplified?

[Josie Jones]

Hi, I am trying to simplify a double summation and was wondering if anyone would be able to help me.

The sum is

$$ \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i a_j $$

Is it possible to simplify it down and maybe lose one of the sigmas?

Thank you in advance :)

 

[mfb]

Yes. Consider $\displaystyle \sum_{i=1}^{n-1} \sum_{j=1}^{i-1} a_i a_j$. You can rearrange that until it looks like your original problem. This, plus the original sum, plus the sum over $a_i^2$ can be expressed much shorter, and that gives you a way to express your sum in a more convenient way as well.

All these steps are easier to follow if you draw a table (i,j).

 

[mathman]

There are no $a_i^2$ terms.

 

[mfb]

There are no $a_i^2$ terms.

Not in the original sum, but if you add these terms (in a suitable way) you can get a nice compact expression. The original sum is then the difference between a nice compact expression and the sum of these squares (with suitable prefactors).

 

[BWV]

so if this $$ \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_i a_j $$ is taking elements from the same vector a of size n and summing the products of adjacent elements, then why can't you write it as $$ \sum_{i=2}^{n} a_i La_i $$ where L is the Lag Operator:

for all t >1 (or a and index i in this case)

 

[mfb]

@BWV: That is a completely different sum.

 

[BWV]

Yes it is, the indexes in the OP are such a mess, I guessed at what perhaps was being attempted. The double sum with one index depending on the other does not make any sense to me

 

[mfb]

The double sum with one index depending on the other does not make any sense to me

Huh? That is very common, and the notation is nothing unusual either. The range of elements to be summed over in the inner sum depends on the index of the outer sum. So what?