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Inner Product Proof - Verify on L2[-1,1]

  1. Feb 5, 2015 #1

    ElijahRockers

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    Gold Member

    1. The problem statement, all variables and given/known data
    This question has two parts, and I did the first part already I think.

    If B = {u1, u2, ..., un} is a basis for V, and
    ##v = \sum_{i=1}^n a_i u_i##
    and ##w = \sum_{i=1}^n b_i u_i##

    Show ##<v,w> = \sum_{i=1}^n a_i b_i^* = b^{*T}a##

    Here's how I did it:

    ##<v,w> = <\sum_{i=1}^n a_i u_i , w> = \sum_{i=1}^n a_i<u_i , w>##
    ## = \sum_{i=1}^n a_i b_i^* <u_i , u_i> = \sum_{i=1}^n a_i b_i^*##
    Thus proved... however in class he mentioned ##a_i = <v,u_i >## for doing this but I'm not sure how... I've tried to examine it but I can't seem to justify it. And I think I did the proof without that, since <v,aw> = a*<v,w>

    Second part of the question, where I'm confused, is, verbatim:

    Verify ##V = L^2 [-1,1]##, where B is the set of orthonormal Legendre polynomials,

    ##p_0 (x) = \frac{1}{\sqrt{2}}##
    ##p_1 (x) = \sqrt{\frac{3}{2}}x##
    ##p_2 (x) = \sqrt{\frac{5}{8}}(3x^2 -1)##

    and v,w are replaced by ##x-x^2## and ##12+x-3x^2##

    2. Relevant equations


    3. The attempt at a solution
    Not really sure where to start... he mentioned a_i = <v, u_i > in class but I don't really feel comfortable with using that here because I don't understand how that's true. ( I feel like it's really simple, and that's why it's bothering me so much) If somebody could point me in the right direction as to why that expression is true, I could probably finish the question..
     
  2. jcsd
  3. Feb 5, 2015 #2

    Ray Vickson

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    Homework Helper

    Write ##u(x) =x-x^2## and ##v(x) = 12 + x - 3x^3## as constant-coefficient linear combinations of ##p_0(x), p_1(x), p_2(x)##. You can do it using your instructor's hint, or you can do it for ##u(x)## the hard way, by getting three equations for ##a_0, a_1,a_2## from the identities ##u(x) = a_0 p_0(x) + a_1 p_1(x) + a_2 p_2(x) \; \forall x##. Do the same type of thing for ##v(x)##. Then, if you want to, you can verify explicitly that ##a_i = \langle u,p_i \rangle##, etc.
     
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