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Math Help - inner product find basis

  1. #1
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    inner product find basis

    Let C[0,\pi] be the complex inner product space of complex continuous functions on [0,\pi], with inner product
     \langle f , g \rangle = \int_0^{\pi} f(x) \overline{g(x)} \, dx.
    Find an orthogonal basis for the subspace of C[0,\pi] spanned by the functions {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 C[0,\pi] be the complex inner product space of complex continuous functions on [0,\pi], with inner product
     \langle f , g \rangle = \int_0^{\pi} f(x) \overline{g(x)} \, dx.
    Find an orthogonal basis for the subspace of C[0,\pi] spanned by the functions {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 v_1=1\,,\,v_2=2ix\,,\,v_3=x^3 , then

    u_i=\frac{v_1}{\|v_1\|} , where of course \|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}}

    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}} =2ix-\frac{2i}{\pi}\cdot \frac{\pi^2}{2}=2ix-\pi i , and now u_2=\frac{w_2}{\|w_2\|}

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

    Tonio
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