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Gram-Schmidt procedure to find orthonormal basis

  1. Sep 19, 2011 #1
    1. The problem statement, all variables and given/known data

    The four functions v0 = 1; v1 = t; v2 = t^2; v3 = t^3 form a basis for the vector space of
    polynomials of degree 3. Apply the Gram-Schmidt procedure to find an orthonormal basis with
    respect to the inner product: < f ; g >= (1/2)[itex]\int[/itex] 1-1 f(t)g(t) dt


    2. Relevant equations

    ui = vi - [itex]\sum[/itex]i-1j <vi, uj>/||vj||2> *vj

    3. The attempt at a solution

    I am not sure that the impact that given inner product integral gives to the question. I don`t even know how to approach this question as well, because I have typically been given vectors of the form (x1, y1, z1) to use gram-schmidt orthonormalization with, not of the given form.
     
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  3. Sep 19, 2011 #2

    Ray Vickson

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    What is stopping you from using the formula in (2) with the inner product as defined in (1)?

    RGV
     
  4. Sep 19, 2011 #3
    How would I go about doing that? I'm a little bit confused about what the relationship between f(t), g(t), vi and ui is in this case. I understand that I can put the integral from part 1 into the inner product for the gram-schmidt orthogonalization in part 2, but what would my f(t) and g(t) represent?
     
  5. Sep 19, 2011 #4

    Ray Vickson

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    They would be whatever two functions whose inner product you want.

    RGV
     
    Last edited: Sep 19, 2011
  6. Sep 19, 2011 #5

    I like Serena

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    Welcome to PF, MellyC! :smile:

    Let's take the first two as examples:

    [tex]u_0 = v_0[/tex]
    [tex]u_1 = v_1 - \sum_{j=0}^{1-1} {<v_1, u_j> \over ||v_j||^2} \cdot v_j[/tex]


    So:
    [tex]u_0 = 1[/tex]
    [tex]u_1 = t - {<t, 1> \over ||1||^2} \cdot 1[/tex]
    with:
    [tex]<t, 1> = {1 \over 2} \int_{-1}^1 t \cdot 1 dt[/tex]
    [tex]||1||^2 = <1, 1> = {1 \over 2} \int_{-1}^1 1 \cdot 1 dt[/tex]


    Can you calculate u1 from this?
    And u2 and u3?
     
  7. Sep 19, 2011 #6

    vela

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    If you really want to, you can look at the polynomials like that. A polynomial f(t) is a linear combination of your given basis vectors:
    [tex]f(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3 = a_0 \vec{v}_0 + a_1 \vec{v}_1 + a_2 \vec{v}_2 + a_3 \vec{v}_3[/tex]so f(t) corresponds to the coordinate vector (a0, a1, a2, a3) and vice versa. You could do your calculations using the 4-tuples (except the inner product since it involves integrating the functions) and then convert your answers back to polynomial form.

    But you should be able to see that scalar multiplication of a 4-tuple is the same thing as the usual scalar multiplication of the polynomial. Similarly, vector addition of the 4-tuples is the same as adding the two polynomials the way you normally do. So there's really no reason to use the 4-tuples instead of the polynomials directly in the calculations. They're just different ways of writing the same thing.
     
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