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Math Help - [SOLVED] Best approximation/least squares/inner product

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Best approximation/least squares/inner product

    Let f^*=\sum_{j=0}^n c_j^* \phi_j be the best approximation with respect to f for the least squares, where { \phi_0,\phi_1,...,\phi_n} is a linear independent set. Show that \parallel f-f^* \parallel={\parallel f \parallel}^2-{\parallel f^* \parallel}^2 where the norm is the one induced by the inner product.
    My work : I've been up to \parallel f-f^* \parallel={\parallel f \parallel}^2-{\parallel f^* \parallel}^2-2<f,f^*>. So if I show that <f,f^*>=0, then I'm done. I'm stuck on this. (I think I must use the fact that the set is linear independent... but how?) Can you help me?
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by arbolis View Post
    Let f^*=\sum_{j=0}^n c_j^* \phi_j be the best approximation with respect to f for the least squares, where { \phi_0,\phi_1,...,\phi_n} is a linear independent set. Show that \parallel f-f^* \parallel={\parallel f \parallel}^2-{\parallel f^* \parallel}^2 where the norm is the one induced by the inner product.
    My work : I've been up to \parallel f-f^* \parallel={\parallel f \parallel}^2-{\parallel f^* \parallel}^2-2<f,f^*>. So if I show that <f,f^*>=0, then I'm done. I'm stuck on this. (I think I must use the fact that the set is linear independent... but how?) Can you help me?
    i don't think if this would help..

    my idea is to show that \parallel f-f^* \parallel \leq {\parallel f \parallel}^2-{\parallel f^* \parallel}^2 and \parallel f-f^* \parallel \geq {\parallel f \parallel}^2-{\parallel f^* \parallel}^2

    you are already done with the first job, just do the second one..
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