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Linear Algebra, Orthonormal question

  1. Mar 1, 2012 #1
    1. The problem statement, all variables and given/known data

    I have this question that I am trying to figure out about orthonormality,I have tried to take a picture of it and put it on here but I can't figure out the url. Anyway I will try and write it out.

    Show that the vector {sin(x),cos(x)} is a basis for the vector space defined by:
    V={asin(x) + bcos(x) l a,b ε ℝ, 0≤ x ≤ pi} using the inner product :
    <f,g>=∫(0 to pi)fgdx, fgεV
    and determine an orthonormal basis.


    2. Relevant equations



    3. The attempt at a solution

    I found the integral of sin(x)cos(x)dx between 0 and pi to be 0.
    This make it orthonormal right as it's the same as the dot product.

    Now I think I have to find out whether <f,f> is = 1 (of unit length)

    so I did integral of sin^2(x)dx between 0 and pi but found pi/2. I also found the same for cos^2(x).

    Does this mean they are not orthonormal? I don't know if it makes a difference that <f,f>=<g,g>.

    Thanks
     
  2. jcsd
  3. Mar 1, 2012 #2

    jbunniii

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    Correct, sin(x) and cos(x) are orthogonal [their inner product with each other is zero] but not orthonormal [their norms are not 1]. There's a very simple modification you can make to sin(x) and cos(x) which will make them orthonormal. Do you see what it is?
     
  4. Mar 1, 2012 #3

    jbunniii

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    P.S. It's clear that V is spanned by sin(x) and cos(x), but the definition of a basis also requires that these functions must be linearly independent. Did you show that?
     
  5. Mar 1, 2012 #4
    What I find helps when you're first working with the notion of 'functions as vectors' is to look back to your basic 3-vectors and think about what these properties mean for them.

    Orthogonal 3-vectors have a.b = 0, orthogonal functions have f.g = 0
    Normalised vectors have a.a = 1, normalised functions have f.f = 1

    Think about how you normalise 3-vectors. Think about how you orthogonalise 3-vectors.
    Here's another little problem you can do which is kinda fun, build up a set of orthogonal polynomials using the gram-schmidt process (these polynomials are the legendre polynomials)
     
  6. Mar 1, 2012 #5
    I think so, I sort of spotted it but wasn't sure. Do I define some sort of new function, lets say h(x), where h(x) = (π^-0.5)sin(x)? Then <h,h> = 1? This is normalization right?
     
    Last edited: Mar 1, 2012
  7. Mar 1, 2012 #6
    I can explain it but don't know how to write this in a neat way. For example if I plug in values for x between 0 and π, lets take 0, then I know that this makes the sin term 0 and cos positive, and if we take pi/2 for example then it's the other way around. So for whatever value is taken between 0 and pi, one of them is a postive and one is 0, therefore a=b=0.

    I however don't know how to express this in proper notation.
     
  8. Mar 1, 2012 #7

    jbunniii

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    Right idea, but I think your scale factor is off. If your original function had <f,f> = pi/2, then you would want to define h = sqrt(2/pi)*f.
     
  9. Mar 1, 2012 #8
    I mean <f,f> = pi sorry, then h=pi^(-0.5)*f would work?
     
  10. Mar 1, 2012 #9

    jbunniii

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    Yes, you have the right idea. You want to show that if

    a cos(x) + b sin(x) = 0 for all x,

    then a and b must be 0.

    So pick some specific values of x that are easy to work with. I suggest x = 0 and x = pi/2.
     
  11. Mar 1, 2012 #10

    jbunniii

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    But is <f,f> = pi?

    Isn't it

    [tex]\int_{0}^{\pi} sin^2(x) dx[/tex]?

    This should work out to pi/2, not pi. And similarly for cos(x).
     
  12. Mar 1, 2012 #11

    Yeh originally I did it from 0 to pi but when I double checked for some reason I made a mistake and did it from -pi to pi.


    Ok so I just need (pi/2)^(-1)?
     
  13. Mar 1, 2012 #12
    Yeh I tried these,

    acos(0)+bsin(0)=0
    Then a must be 0.

    acos(pi/2)+bsin(pi/2)=0
    then b must be 0.

    Is it enough to write this, or is there some sort of other form I can generalize it for?
     
  14. Mar 1, 2012 #13

    jbunniii

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    The square root of that.
     
  15. Mar 1, 2012 #14

    jbunniii

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    That's enough. If the linear combination equals zero, it has to equal zero at every value of x, including the two you chose (0 and pi/2). Those two alone are enough to imply that a = b = 0. You could have picked almost any pair of values of x, as long as they resulted in a solvable system of two equations with two unknowns. 0 and pi/2 just make the math easier.
     
  16. Mar 1, 2012 #15
    Thanks jbunniii,

    I understand all of it, and have managed to complete 2 similar questions now.

    My only question is, why is it the square root? Surely if <f,f> = pi/2 then we need 2/pi to make this 1?
     
  17. Mar 2, 2012 #16

    jbunniii

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    No, suppose you set [itex]h = \alpha f[/itex]. Then

    [tex]<h,h> = <\alpha f, \alpha f> = \alpha^2 <f,f>[/tex]

    You want this equal to 1, so

    [tex]\alpha^2 = 1/<f,f>[/tex]

    and hence

    [tex]\alpha = \sqrt{1/<f,f>}[/tex]
     
  18. Mar 2, 2012 #17
    Yeh I was thinking just at the end not the fact that they are going to multiply each other. Thanks for all your help Jbunnii
     
    Last edited: Mar 2, 2012
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