How can I generalize the squaring of any sum of variables?

[Sigma057]

Hello, I am a recently graduated high school senior going off to college soon, and over the summer I have spend some of my free time experimenting with different problems in mathematics. This is one that I have spent many days on, and have come to a dead end.

Like the title says, I am attempting to generalize the squaring of any sum of variables

$$(x_1+ \dots + x_n)^2$$

So far I've just experimented with multiplying it out with different values of n and trying to spot a pattern.

For n = 3 it comes out to be
$$x^2_1 +x^2_2 +x^2_3 + 2(x_1 x_2 + x_1 x_3 + x_2 x_3)$$

The pattern seems to be the squares of the variables + 2 times all the combinations of 2 variables multiplied together.

I thought higher values of n would help me get a general pattern so I chose n = 6
$$(x_1 + \dots + x_6)^2 = x^2_1 + x^2_2 + x^2_3 + x^2_4 + x^2_5 + x^2_6 + 2(x_1 x_2 + x_1 x_3 + x_1 x_4 + x_1 x_5 + x_1 x_6 + x_2 x_3 + x_2 x_4 + x_2 x_5 + x_2 x_6 + x_3 x_4 + x_3 x_5 + x_3 x_6 + x_4 x_5 + x_4 x_6 + x_5 x_6)$$

Since there are so many terms to be added I thought using Sigma notation would help. The same formula above in Sigma notation is

$$( \sum_{i=1}^6 x_i )^2 = \sum_{i=1}^6 x^2_i + 2( \sum_{i=2}^6 x_1 x_i + \sum_{i=3}^6 x_2 x_i +\sum_{i=4}^6 x_3 x_i + \sum_{i=5}^6 x_4 x_i + x_5 x_6 )$$

This is obviously not a very compact formula, and I was wondering if there is a way to combine the sums of different indices. Or, if there is a better way to go about this.

 

[Citan Uzuki]

$$\left( \sum_{i=1}^n x_i \right)^{2} = \sum_{i=1}^{n}x_{i}^{2} + 2 \sum_{1 \leq i < j \leq n}x_{i} x_{j}$$

Does that help?

 

[Sigma057]

$$\left( \sum_{i=1}^n x_i \right)^{2} = \sum_{i=1}^{n}x_{i}^{2} + 2 \sum_{1 \leq i < j \leq n}x_{i} x_{j}$$

Does that help?

I've never used a summation with 2 indices without being a double sum. What exactly does the second sum mean?

 

[Citan Uzuki]

It means the sum over all ordered pairs (i, j) where i<j. You could rewrite it using a double sum as follows:

$$\sum_{i=1}^{n} x_i^2 + 2 \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} x_i x_j$$

Or more directly, without separating out the cross terms from the ones with i=j as:

$$\sum_{i=1}^{n} \sum_{j=1}^{n} x_i x_j$$

 

[daveyp225]

A quick way to realize this simplification might be:

$$(x_1+x_2 +\cdots + x_n)(x_1+x_2 +\cdots + x_n) = x_1(x_1+x_2 +\cdots + x_n) + x_2(x_1+x_2 +\cdots + x_n) + \cdots+ x_n(x_1+x_2 +\cdots + x_n)$$

The $k^{th}$ term above contains an $x_k^2$ and $x_ix_k$ for every $i \neq k$. So the $k^{th}$ and the $i^{th}$ are the only terms which contain $x_kx_i=x_ix_k$ giving you two copies of every non-square.

If your ring were not commutative, the general result would be

$$(x_1+x_2+\cdots + x_n)^2 = \sum_{i\le n} x_i^2 + \sum_{i \neq j} x_ix_j$$

Which looks (maybe deceptively) less complicated. Of course this applies to the commutative case as well.

 

[disregardthat]

And now, (x_1 + x_2 + ... + x_n)^m ...

 

[Sigma057]

And now, (x_1 + x_2 + ... + x_n)^m ...

Yes! That will be my next step.

 

[daveyp225]

Yes! That will be my next step.

The notation might get very messy. I suggest you try the third power to see how it relates or comes from to the second power (then, if needed, see how the 4th power relates to the third). Then try to generalize any power.