Thread: symmetry expansion of a polynomial.

1. symmetry expansion of a polynomial.

hi everyone,

I have this equation which has sigma notation. I need to expand the sigma into factors and solve for x.

$\underset{abc}{\Sigma}(x-a)[(x-a)^2-(x-b)(x-c)]=0$

my question is did I get the symmetry expansion right? I want to make sure before I try to solve for x. also, in general how do you think about this? I think you have 3 possible combinations in first factor. each of those has 3 combinations of a,b,c. 3+3+3=9 terms in total.

$\\(x-a)[(x-a)^2-(x-b)(x-c)]+\\(x-a)[(x-b)^2-(x-c)(x-a)]+\\(x-a)[(x-c)^2-(x-a)(x-b)]+\\(x-b)[(x-a)^2-(x-b)(x-c)]+\\(x-b)[(x-b)^2-(x-c)(x-a)]+\\(x-b)[(x-c)^2-(x-a)(x-b)]+\\(x-c)[(x-a)^2-(x-b)(x-c)]+\\(x-c)[(x-b)^2-(x-c)(x-a)]+\\(x-c)[(x-c)^2-(x-a)(x-b)]=0$

is this the total expansion?

2. Hello, skoker!

$\sum_{abc}(x-a)[(x-a)^2-(x-b)(x-c)]\:=\:0$
.
What is this?

I don't understand how that $abc$ can be interpreted to mean:
. . replace $a,b,c$ with every permutation of $\{a,b,c\}.$

3. @Soroban

this is notation for absolute-symmetric sum and cyclic-symmetric sum. where you have terms of the same type having the same coefficient. it shows the contraction of the sum of terms of the same type.

$\underset{ab}{\Sigma}a^3=a^3+b^3$ also $\underset{abc}{\Sigma}a^3=a^3+b^3+c^3$

you only need to put abc under sigma if it is not obvious what variables to use.

$(a^2+b^2+c^2)(a+b+c)=\Sigma a^3+\Sigma a^2 b \rightarrow \underset{abc}{\Sigma}a^2 \underset{abc}{\Sigma}a= \underset{abc}{\Sigma}a^3+\underset{abc}{\Sigma}a^ 2 b$

a single term is understood to be absolute-symmetric. to denote cyclic use a letter like S. you can also use 'sym' and 'cyc'.

$\Sigma}a^2 b=\underset{sym}{\Sigma}a^2 b=a^2 b+b^2 c+c^2 a+b^2 a+c^2 b+a^2 c$

$Sa^2 b=\underset{cyc}{\Sigma}a^2 b=a^2 b+b^2 c+c^2 a$

if there is more then one term in the group then cyclic symmetry is implied not absolute.

so the cyclic expansion of the expression is.

$\\(x-a)[(x-a)^2-(x-b)(x-c)]\\(x-c)[(x-c)^2-(x-a)(x-b)]\\(x-b)[(x-b)^2-(x-c)(x-a)]\\$

I notice that what I did was not correct in the first case. I did to many cycles, I did cycle combinations according to brackets. but the crazy part is that after you expand and collect like terms both versions are IDENTICAL??? I try to think about why this is the case. It has something to do with fact that the first factor has full symmetry and the last 3 have cyclic. so maybe that cancels out somehow through one complete cycle.

first case expansion:

$\Sigma a^3+3\Sigma a^2 x-3\Sigma abx+3abc + 2\Sigma a^2 b -2\Sigma a^2 b$

second case expansion:

$\Sigma a^3+3\Sigma a^2 x-3\Sigma abx+3abc+ 2\Sigma ax^2 -2\Sigma ax^2$

4. I think this is solution for x. I use the sigma and the problem becomes easy to work through.

first the cyclic-expansion.

$\underset{abc}{\Sigma}(x-a)[(x-a)^2-(x-b)(x-c)]=$

$\\(x-a)[(x-a)^2-(x-b)(x-c)]+\\(x-c)[(x-c)^2-(x-a)(x-b)]+\\(x-b)[(x-b)^2-(x-c)(x-a)]=$

$-a^3-b^3-c^3+3a^2 x+3b^2 x+3c^2 x-3abx-3cax-3bcx+3abc=$

$-\Sigma a^3+3\Sigma a^2x-3\Sigma abx+3abc.$

then solve for x.

\begin{aligned}-\Sigma a^3+3\Sigma a^2x-3\Sigma abx+3abc & =0\\3\Sigma a^2x-3\Sigma abx & =\Sigma a^3-3abc\\3x(\Sigma a^2-\Sigma ab) & =\Sigma a(\Sigma a^2-\Sigma ab) \hspace*{10 pt}\bullet (a+b+c)(a^2 +b^2 +c^2 -ab-ca-bc)=\Sigma a^3 -3abc\\x & =\frac{\Sigma a}{3}\\x & =\frac{a+b+c}{3}.\\\end{aligned}

5. so I do a check using sigma again to lean to do calculations mentally. each term is absolute-symmetric and also heterogeneous so the product is also absolute-symmetric and heterogeneous. I realize that abc term has 3 sym and could have sigma also but that seems more confusing on paper.

check:

\begin{aligned}0&= -\Sigma a^3 + 3\Sigma a^2 x - 3\Sigma abx +3abc\\0&= -\Sigma a^3 + 3\Sigma a^2 \Big(\frac{\Sigma a}{3} \Big) - 3\Sigma ab\Big(\frac{\Sigma a}{3} \Big) +3abc\\0&=-\Sigma a^3 + \Sigma a^2 \Sigma a - \Sigma ab \Sigma a +3abc\\0&=-\Sigma a^3 + \Sigma a^3+ \Sigma a^2b -\Sigma a^2b -3abc +3abc\\0&=0.\end{aligned}

so we get identity 0, answer is correct.