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Normalising orthogonal basis

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data
    I have used the gram schmidt process to find an orthogonal basis for [itex]{1,t,t^2}[/itex]
    which is
    [itex]
    (1,x,x^2 - \frac{2}{3})
    [/itex]

    How to i normalize these

    2. Relevant equations

    [itex]e_1=\frac{u_1}{|u_1|}[/itex]

    3. The attempt at a solution

    [itex]e_1=\frac{1}{\sqrt{\int_{-1}^{1}f(1)g(1)}}=\frac{1}{\sqrt{2}}[/itex]


    [itex]e_2=\frac{x}{\sqrt{\int_{-1}^{1}f(x)g(x)}}=\frac{x}{\sqrt{\frac{2}{3}}}[/itex]

    [itex]e_3=\frac{x}{\sqrt{\int_{-1}^{1}f(x^2-3)g(x^2-3)}}=\frac{x^2}{\sqrt{\frac{-62}{45}}}[/itex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 26, 2009 #2
    Just divide each by its norm.
     
  4. Oct 26, 2009 #3

    Mark44

    Staff: Mentor

    There is a problem with your third function. It is orthogonal to x, but not to 1. Each function has to be orthogonal to each other function in your set.
    [tex]\int_{-1}^1 1 (x^2 - 2/3) dx~=~\left[\frac{x^3}{3} - (2/3)x\right]_{-1}^1~=~ -2/3[/tex]

    Also, is the function x^2 - 2/3 or x^2 - 3? You are using both in your calculations.
     
  5. Oct 26, 2009 #4

    Mark44

    Staff: Mentor

    One other thing. The norm of a function is the square root of the inner product of it and itself.
    You shouldn't have f(x) and g(x) in there for <x, x>, since f(x) = g(x) = x. It should be like this:
    [itex]e_2=\frac{x}{\sqrt{\int_{-1}^{1}x*xdx}}= ...[/itex]
     
  6. Oct 26, 2009 #5
    thanks your right my u3 should be [itex]x^2-\frac{1}{3}[/itex]
    which makes
    [itex]e_3=\frac{x^{2}-\frac{1}{3}}{\sqrt{\int_{-1}^{1}x^{2}-\frac{1}{3}*x^{2}-\frac{1}{3}dx}}= ...[/itex]

    = [itex]\frac{x^2-\frac{1}{3}}{\sqrt{frac{8}{45}}}[/itex]

    hows that look

    The innner product of [itex]x^2-3 and \frac{1}{\sqrt{2}} = 0[/itex] so they're orthogonal
     
  7. Oct 26, 2009 #6

    Mark44

    Staff: Mentor

    Much better.

    For the quantity in the radical, you left off the \ before frac in your LaTeX code. That's why it looks like it does. Should look like this:
    [itex]\frac{x^2-\frac{1}{3}}{\sqrt{\frac{8}{45}}}[/itex]
     
  8. Oct 27, 2009 #7
    Thanks for your help guys much appreciated
     
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