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Math Help - how did they do this transition..

  1. #1
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    how did they do this transition..

    http://i36.tinypic.com/lw3uh.jpg

    i would have transform each norm into inner product form
    and use the distributive law.

    but here they just do some magic
    put out thr Re part

    i dont know whats the Re part in that context

    what method they use here?

    as if they say
    ||f+g||^2 =(||f||+||g||)^2

    thats not true its a wrong way of the triangle innequality
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    http://i36.tinypic.com/lw3uh.jpg

    i would have transform each norm into inner product form
    and use the distributive law.

    That's what they did but in a long, sloppy way:

    ||f+if||-||f-if||=|1+i|\cdot||f||-|1-i|\cdot||f||=0\,,\,\,since\,\,||kf||=|k|\cdot||f||  \,,\,\,k\in \mathbb{C}[/color]

    but here they just do some magic
    put out thr Re part

    i dont know whats the Re part in that context

    ||f+if||=<f,f>-i<f,f>+i<f,f>+<f,f> since\,\,<x,ky>=\overline{k}<x,y> , but for some reason I can't understand, instead of cancelling the

    two central summands, they thought it'd be a good idea to write

    i<f,f>-i<f,f>=-<f,if>-\overline{<f,if>}=-2Re(<f,if>) , which seems a pretty goofy and weird thing to do, though the final result still is right.

    Tonio

    what method they use here?

    as if they say
    ||f+g||^2 =(||f||+||g||)^2

    thats not true its a wrong way of the triangle innequality
    .
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  3. #3
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    cant understad this line
    <br />
i<f,f>-i<f,f>=-<f,if>-\overline{<f,if>}=-2Re(<f,if>) <br />


    <br />
i<f,f>-i<f,f>=0<br />

    i do understand this step
    <br />
-<f,if>-\overline{<f,if>}<br />
    but why
    <br />
-<f,if>-\overline{<f,if>}=-2Re(<f,if>)<br />

    ?
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    cant understad this line
    <br />
i<f,f>-i<f,f>=-<f,if>-\overline{<f,if>}=-2Re(<f,if>) <br />


    <br />
i<f,f>-i<f,f>=0<br />

    i do understand this step
    <br />
-<f,if>-\overline{<f,if>}<br />
    but why
    <br />
-<f,if>-\overline{<f,if>}=-2Re(<f,if>)<br />

    ?

    Hmmm...I think you're rushing to write back without properly thinking things over: \forall z\in \mathbb{C}\,,\,\,z+\overline{z}=2Re(z) , as you can easily check if, for example, you write z=x+iy\,,\,\,z\,,\,y\in \mathbb{R} , and then put z=-<f,if>

    Tonio
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  5. #5
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    yes i understand now thanks
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