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Math Help - Help with a proof.

  1. #1
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    Help with a proof.

    Hi, I am having trouble with this proof, I am wondering what to do because I keep getting the wrong answer.

    I want to prove

    ||x-y|| = ||x|| ||y|| ||x bar - y bar||

    where

    x and y are vectors in Rn

    and u bar is defined by u/(||u||^2)


    The way i started it was expanding the right side completely by using the norm formula and then simplify it into making it equal to the left, but I end up with the wrong answer and a really messy calculation. I was wondering if there was a simpler way to do this.
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  2. #2
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    Re: Help with a proof.

    Quote Originally Posted by Kuma View Post
    Hi, I am having trouble with this proof, I am wondering what to do because I keep getting the wrong answer.
    I want to prove
    ||x-y|| = ||x|| ||y|| ||x bar - y bar||
    where
    x and y are vectors in Rn
    and u bar is defined by u/(||u||^2)
    It seems to me as if there is something missing here.
    You write that \overline{u}=\frac{u}{\|u\|^2}

    You are asked to prove that
    \|x-y\|=\|x\|\|y\|\|\overline{x}-\overline{y}\|.

    Is that correct? What more is there to \|x-y\|~?
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  3. #3
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    Re: Help with a proof.

    Well this is what the question asks exactly

    Given a nonzero vector u E R^n, put u bar as:

    (1/||u||^2)*u

    and the rest is the same, equality you put is correct, that is what it asks me to prove.

    I figured that (1/||u||^2)*u would be the same as u/||u||^2, the way the question writes it is (1/||u||^2)*u.

    Any ideas as to what is missing? When I try to do the proof by going the long hard way and expanding the norm of x bar - y bar, I get close to the left side but its not quite there after simplifying everything.
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  4. #4
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    Re: Help with a proof.

    Quote Originally Posted by Kuma View Post
    When I try to do the proof by going the long hard way and expanding the norm of x bar - y bar,
    What exactly does that mean?
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  5. #5
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    Re: Help with a proof.

    Quote Originally Posted by Plato View Post
    What exactly does that mean?
    Ok so this is what i did for the norm of x bar - y bar

    if x = (x1,...,xn) and y = (y1,...,yn)

    then i have

    ||(x1,...,xn)/[sqrt(x1^2+...+xn^2)]^2 - (y1,...,yn)/[sqrt(y1^2+...+yn^2)]^2||

    im just going to denote sqrt(x1^2+...+xn^2) = c and sqrt(y1^2+...+yn^2) = d

    =sqrt[ (x1/(c^2) - y1/(d^2))^2 +...+(xn/(c^2) - yn/(d^2))^2]

    and I went from there onward by simplifying. Is that the right way to do it?
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  6. #6
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    Re: Help with a proof.

    ||x||||y||||\overline{x}-\overline{y}|| = ||x||||y||\sqrt{(\overline{x}-\overline{y})\cdot(\overline{x}-\overline{y})}

    =||x||||y||\sqrt{(\overline{x}\cdot\overline{x}) - 2(\overline{x} \cdot\overline{y}) + (\overline{y}\cdot\overline{y})}

    =||x||||y||\sqrt{(\frac{x}{||x||^2}\cdot \frac{x}{||x||^2}) - 2(\frac{x}{||x||^2}\cdot \frac{y}{||y||^2}) + (\frac{y}{||y||^2} \cdot \frac{y}{||y||^2})}

    =||x||||y||\sqrt{\frac{1}{||x||^2} - 2(\frac{x\cdot y}{||x||^2||y||^2}) + \frac{1}{||y||^2}}.

    =||x||||y||\sqrt{\frac{||y||^2 - 2(x \cdot y) + ||x||^2}{||x||^2||y||^2}}

    =\sqrt{(||x||^2 - 2(x \cdot y) + ||y||^2}

    =\sqrt{(x \cdot x) - 2(x \cdot y) + (y \cdot y)} = \sqrt{(x-y)\cdot(x-y)}

    =||x-y||
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