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Showing that the following are equivalent

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data

    I had this posted under a different question a while back and didn't get any responses, so I thought I'd rephrase it. I've reduced it to what I think I'm supposed to show. (Here's the old post: https://www.physicsforums.com/showthread.php?p=4260206#post4260206 ... Disregard the actual coefficients, I fudged those a bit.)

    2. Relevant equations

    3. The attempt at a solution
    I have to show the following:
    ##||x||^2 = \frac{2}{3}|x_1|^2+\frac{1}{6}|x_1-\sqrt3x_2|^2+\frac{1}{6}|x_1+\sqrt3x_2|^2##

    ##x\in\mathbb{R}^2##, so ##x=(x_1,x_2)##, and the left side can be rewritten so that I have
    ##|x_1|^2+|x_2|^2 = \frac{2}{3}|x_1|^2+\frac{1}{6}|x_1-\sqrt3x_2|^2+\frac{1}{6}|x_1+\sqrt3x_2|^2##

    Is there any way to do this?
     
  2. jcsd
  3. Feb 8, 2013 #2

    SammyS

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    |a + b|2 = (a + b)2

    The rest is pretty basic algebra.

    Merely expand the right hand side & collect terms.
     
  4. Feb 8, 2013 #3
    Is it really that simple? I was getting the impression I couldn't do that. I thought for some reason that the sign of x1 or x2 would make a difference. Thanks a lot!
     
  5. Feb 9, 2013 #4

    SammyS

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    If a+b ≥ 0 , then it's pretty obvious that |a+b|2 = (a+b)2 , since |a+b| = a+b .

    If a+b < 0 , then |a+b| = -(a+b), so that |a+b|2 = (-(a+b))2. But that's obviously equal to (a+b)2.
     
  6. Feb 9, 2013 #5
    Oh I see it now, thanks! I was thinking that the sign of x1 and x2 would have some effect on the sum. Thanks again.
     
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