Results 1 to 5 of 5

Math Help - symmetry expansion of a polynomial.

  1. #1
    Member
    Joined
    Oct 2010
    From
    Heart of Winter
    Posts
    115
    Thanks
    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,550
    Thanks
    542
    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\}.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2010
    From
    Heart of Winter
    Posts
    115
    Thanks
    1
    @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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2010
    From
    Heart of Winter
    Posts
    115
    Thanks
    1
    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}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2010
    From
    Heart of Winter
    Posts
    115
    Thanks
    1
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: February 28th 2011, 04:07 AM
  2. Replies: 6
    Last Post: May 1st 2009, 11:37 AM
  3. Polynomial Expansion Using Pascals Triangle
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 3rd 2008, 08:29 AM
  4. Symmetry
    Posted in the Geometry Forum
    Replies: 2
    Last Post: May 8th 2006, 01:31 PM
  5. polynomial expansion
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 30th 2005, 07:59 AM

Search Tags


/mathhelpforum @mathhelpforum