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Thread: inner product find basis

  1. #1
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    inner product find basis

    Let $\displaystyle C[0,\pi]$ be the complex inner product space of complex continuous functions on $\displaystyle [0,\pi]$, with inner product
    $\displaystyle \langle f , g \rangle = \int_0^{\pi} f(x) \overline{g(x)} \, dx. $
    Find an orthogonal basis for the subspace of $\displaystyle C[0,\pi]$ spanned by the functions $\displaystyle {1,2ix,x^3}$

    I think I have to used Gram-Schmidt method but I am not sure where to start. Can you please show me the general method and I will then be able to solve it.
    Many thanks
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  2. #2
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    Quote Originally Posted by charikaar View Post
    Let $\displaystyle C[0,\pi]$ be the complex inner product space of complex continuous functions on $\displaystyle [0,\pi]$, with inner product
    $\displaystyle \langle f , g \rangle = \int_0^{\pi} f(x) \overline{g(x)} \, dx. $
    Find an orthogonal basis for the subspace of $\displaystyle C[0,\pi]$ spanned by the functions $\displaystyle {1,2ix,x^3}$

    I think I have to used Gram-Schmidt method but I am not sure where to start. Can you please show me the general method and I will then be able to solve it.
    Many thanks

    Apply EXACTLY Gram-Schmidt Method's proof...what's the problem?

    For example, if $\displaystyle v_1=1\,,\,v_2=2ix\,,\,v_3=x^3$ , then

    $\displaystyle u_i=\frac{v_1}{\|v_1\|}$ , where of course $\displaystyle \|v_1\|=\sqrt{<v_1,v_1>}=\sqrt{\int\limits_0^\pi 1\cdot 1\,dx}=\sqrt{\pi}\Longrightarrow u_1=\frac{1}{\sqrt{\pi}}$

    $\displaystyle w_2=v_1-<v_1,u_1>u_1=2ix-\int\limits_0^\pi \frac{2ix}{\sqrt{\pi}}dx\cdot\frac{1}{\sqrt{\pi}}$ $\displaystyle =2ix-\frac{2i}{\pi}\cdot \frac{\pi^2}{2}=2ix-\pi i$ , and now $\displaystyle u_2=\frac{w_2}{\|w_2\|}$

    Finally, evaluate $\displaystyle w_3=v_2-<v_2,u_2>u_2-<v_2,u_1>u_1$ and then $\displaystyle u_3=\frac{w_3}{\|w_3\|}$ , and then $\displaystyle u_1,u_2,u_3$ is your orthonormal basis.

    Tonio
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