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Math Help - Orthogonal polynomials and inner product space

  1. #1
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    Orthogonal polynomials and inner product space

    Say we have a set {v1, v2, ..., vk, ...} of orthogonal polynomials in some real inner product space < , >, and assume they're normalised to be monic. If q is a polynomial with deg(q) = k, how would I show that

    || vk || <= || q ||, for all q?

    (||x|| = <x,x>^(1/2).)
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  2. #2
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    Quote Originally Posted by alice34 View Post
    Say we have a set {v1, v2, ..., vk, ...} of orthogonal polynomials in some real inner product space < , >, and assume they're normalised to be monic. If q is a polynomial with deg(q) = k, how would I show that

    || vk || <= || q ||, for all q?

    (||x|| = <x,x>^(1/2).)
    the problem in this form is just non-sense! first of all we must have \deg v_k = k and i think we also need a v_0. we also don't need them to be monic. finally if they are monic, then the absolute

    value of the leading coefficient of q should be at least 1 because otherwise q=\frac{1}{2}v_k would be a trivial counter-example.

    anyway, your problem is a quick result of this simple fact that q is a linear combinaton of v_j, \ 0 \leq j \leq k: if q=\sum_{j=0}^kc_jv_j, then \langle q, q \rangle = \sum_{j=0}^k c_j^2 \langle v_j, v_j \rangle \geq c_k^2 \langle v_k,v_k \rangle \geq \langle v_k,v_k \rangle.
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