Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a following problem

When we want to write $a^2 + b^2$ in terms of $(a \pm b)^2$ we can do it like that $$a^2 +b^2 = \frac{(a+b)^2}{2} + \frac{(a-b)^2}{2}.$$

Can we do anything similar for $a_1^2 + a_2^2 + \ldots + a_n^2$ ? I can add the assumption that all $a_i$ are positive numbers. I mean to express this as combination of their sums and differences. I know that this question is a little bit naive but I'm curious whether it has an easy answer.

share|cite|improve this question
Is sunflower's answer enough? – Amr Dec 11 '12 at 15:04
up vote 3 down vote accepted

Yes. You have to sum over all of the possibilities of $a\pm b\pm c$: $$4(a^2+b^2+c^2)=(a+b+c)^2+(a+b-c)^2+(a-b+c)^2+(a-b-c)^2$$

This can be extended to n factors by:

$$\sum_{k=1}^n a_k^2=\sum_{\alpha=(1,-1,...,-1)\; |a_i|=1}^{(1,...,1)}\frac{\big(\sum_{i=1}^{n}\alpha_ia_i\big)^2}{2^{n-1}}$$

($\alpha$ is a multiindex with values that are either -1 or 1, except the first that is always 1)

share|cite|improve this answer

Another way would be:

$$\frac{\sum_{x\in \{-1,1\}^n}(x_1a_1+x_2a_2+....+x_na_n)^2}{2^n}$$

Where $x_i$ is the ith component of the vector $x$

share|cite|improve this answer
I think this is correct but I still need to check. Unfortunately I have to leave now. I will check it later – Amr Dec 11 '12 at 15:09
nice way to write it too ^.^ I wrote an answer using multiindices... this is probably nicer. But should it not be $2^{n-1}$? – CBenni Dec 11 '12 at 15:12
No. It should be $2^n$ – Amr Dec 11 '12 at 15:19
Try the case when $n=2$ or 1 – Amr Dec 11 '12 at 15:20
nah, sorry. I was looking at the solutions for n=2 or 3 where the first index is 1. (These work just fine too). For those, we have only half the terms and thus, only $2^{n-1}$ but you are correct. – CBenni Dec 11 '12 at 15:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.