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Math Help - Inner Products.

  1. #1
    Junior Member pearlyc's Avatar
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    Inner Products.

    For V = C[0,1] with inner product <f,g> = ∫ f(x)g(x)dx [For limits 0 to 1] if f(0) = 0 = f(1) show that f'(x) and f(x) are orthogonal.

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  2. #2
    Super Member PaulRS's Avatar
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    <br />
\left\langle {f,f'} \right\rangle  = \int_0^1 {f\left( x \right) \cdot f'\left( x \right)dx} <br />

    Since: <br />
{\textstyle{1 \over 2}} \cdot \left( {f^2 \left( x \right)} \right)^\prime   = f\left( x \right)f'\left( x \right)<br />
it follows that: <br />
\int_0^1 {f\left( x \right) \cdot f'\left( x \right)dx}  = \left. {{\textstyle{{f^2 \left( x \right)} \over 2}}} \right|_0^1  = {\textstyle{{f^2 \left( 1 \right) - f^2 \left( 0 \right)} \over 2}} = 0<br />

    Thus: <br />
\left\langle {f,f'} \right\rangle  =0<br />
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  3. #3
    Junior Member pearlyc's Avatar
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    May I know how did you come to this part :

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  4. #4
    Super Member PaulRS's Avatar
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    Quote Originally Posted by pearlyc View Post
    May I know how did you come to this part :

    <br />
f^2 \left( x \right) = f\left( x \right) \cdot f\left( x \right)<br />

    and now we apply the product rule: <br />
\left[ {f\left( x \right) \cdot f\left( x \right)} \right]^\prime   = f'\left( x \right) \cdot f\left( x \right) + f\left( x \right) \cdot f'\left( x \right)<br />
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  5. #5
    Junior Member pearlyc's Avatar
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    Oh right!

    Thank you.
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