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

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The discussion focuses on generalizing the squaring of a sum of variables, specifically the expression (x_1 + ... + x_n)^2. The user has identified a pattern for n=3 and n=6, noting that it consists of the squares of the variables plus twice the sum of all combinations of two variables multiplied together. They explore using Sigma notation to express this but find it cumbersome and seek a more compact formula. A suggestion is made to rewrite the sum using double summation to clarify the terms involved. The conversation concludes with encouragement to explore higher powers to further understand the generalization.
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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.
 
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\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?
 
Citan Uzuki said:
\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?
 
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
 
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) +<br /> 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.
 
And now, (x_1 + x_2 + ... + x_n)^m ...
 
disregardthat said:
And now, (x_1 + x_2 + ... + x_n)^m ...

Yes! That will be my next step.
 
Sigma057 said:
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.
 
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