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Math Help - does it defines inner product..

  1. #1
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    does it defines inner product..

    C[0,2]\\
    f:[0,2]->c\\
    <f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2)  \overline{g(2)}

    i dont know how to solve the complement.
    i tried to prove there easier version without the complement first.

    <f,g>=f(0)g(0)+f(1)g(1)+f(2)g(2)\\
    A.<g,f>=g(0)f(0)+g(1)f(1)+g(2)f(2)=f(0)g(0)+f(1)g( 1)+f(2)g(2)=<f,g>
    B.x<f,g>=x(f(0)g(0)+f(1)g(1)+f(2)g(2))=xf(0)g(0)+x f(1)g(1)+xf(2)g(2)=<xf,g>=<f,xg>

    C.<x+y,g>=(x(0)+y(0))g(0)+(x(1)+y(1))g(1)+(x(2)+y( 2))g(2)

    step d:
    <x,x>=x(0)x(0)+x(1)x(1)+x(2)x(2)

    i dont know how to prove that
    <x,x> greater or equal 0
    i dont know the values of x
    ??
    Last edited by transgalactic; October 1st 2009 at 04:23 AM.
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    C[0,2]\\
    f:[0,2]->c\\
    <f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2)  \overline{g(2)}

    i dont know how to solve the complement.
    i tried to prove there easier version without the complement first.

    <f,g>=f(0)g(0)+f(1)g(1)+f(2)g(2)\\
    A.<g,f>=g(0)f(0)+g(1)f(1)+g(2)f(2)=f(0)g(0)+f(1)g( 1)+f(2)g(2)=<f,g>
    B.x<f,g>=x(f(0)g(0)+f(1)g(1)+f(2)g(2))=xf(0)g(0)+x f(1)g(1)+xf(2)g(2)=<xf,g>=<f,xg>

    C.<x+y,g>=(x(0)+y(0))g(0)+(x(1)+y(1))g(1)+(x(2)+y( 2))g(2)

    step d:
    <x,x>=x(0)x(0)+x(1)x(1)+x(2)x(2)

    i dont know how to prove that
    <x,x> greater or equal 0
    i dont know the values of x
    ??


    If we denote the conjugate of a complex number z by z', then you have:

    for a vector x (i.e., x is a function in C[0,2]):

    <x,x> = x(0)x(0)' + x(1)x(1)' + x(2)x(2)' = |x(0)|^2 + |x(1)|^2 + |x(2)|^2

    since zz' = |z|^2 for any complex number z, and thus you clearly have

    <x,x> >= 0 .

    Tonio
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  3. #3
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    i tried to prove step 2:

    <f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2)  \overline{g(2)}\\
    k<f,g>=kf(0)\overline{g(0)}+kf(1)\overline{g(1)}+k  f(2)\overline{g(2)}=<kf,g>\\
    <f,kg>=kf(0)\overline{g(0)}+kf(1)\overline{g(1)}+k  f(2)\overline{g(2)}?=\overline{k}<f,g>

    i dont know how in the end it proves it as
    \overline{k}<f,g>
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    i tried to prove step 2:

    <f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2)  \overline{g(2)}\\
    k<f,g>=kf(0)\overline{g(0)}+kf(1)\overline{g(1)}+k  f(2)\overline{g(2)}=<kf,g>\\
    <f,kg>=kf(0)\overline{g(0)}+kf(1)\overline{g(1)}+k  f(2)\overline{g(2)}?=\overline{k}<f,g>

    i dont know how in the end it proves it as
    \overline{k}<f,g>

    Remember that z'' = z and (Zw)' = z'w' (I use z' to denote complex conjugate), so:

    <f, kg> = f(0)(kg(0))' + f(a) (kg(1))' + f(2)(hg(2))' = f(0)k'g(0)' + f(1)k'g(1)' + f(2)k'g(2)' = k'(f(0)g(0)' + f(1)g(1)' + f(2)g(2)' = k'<f,g>

    Tonio
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  5. #5
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    thanks
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