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Fourier Series Analysis

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that provided that m and n are arbitrary integers, the two functions f(t) = sin nt and g(t) = cos mt are orthogonal over the interval (0,2[tex]\pi[/tex]). Explain the significance of this result in Fourier series analysis. Hint: you may find the following trigonometric identity useful 2sin a cos b = sin(a+b) - sin (a-b)


    2. Relevant equations



    3. The attempt at a solution

    [tex]\int[/tex] f(t) g(t) dt
    [tex]\int[/tex] sin nt cos mt dt
    [tex]\int[/tex] 0.5( sin(n+m)t - sin(n-m)t)dt
    0.5[ [tex]\frac{cos(n+m)t}{n+m}[/tex] - [tex]\frac{cos(n-m)t}{n-m}[/tex] ]

    put (0,2[tex]\pi[/tex]) into (n,m) respectively (sorry not good with code thing).

    After subistion the final answer end up being

    t over 2[tex]\pi[/tex]

    this is my working out so far but what should I write about it under fourier analysis
     
  2. jcsd
  3. Aug 25, 2009 #2

    Dick

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    You don't put (0,2*pi) into (n,m). You set t=2*pi and t=0 into the anti derivative and subtract the results. There's no t in the final answer. What's cos(k*2*pi) where k is an integer?
     
  4. Aug 25, 2009 #3
    so in that case does it become

    [ cos 2*pi*(n+m) / n + m ] - [ cos 2*pi*(n-m) / n - m ]

    ?
     
  5. Aug 25, 2009 #4

    Dick

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    It becomes that minus [ cos 0*(n+m) / n + m ] - [ cos 0*(n-m) / n - m ]. Haven't you done definite integrals before?
     
  6. Aug 25, 2009 #5
    yeah sorry I made a mistake cancelling that because cos 0 = 1 and forget that the denominator where different

    so it becomes

    [ cos 2*pi*(n+m) / n + m ] - [ cos 2*pi*(n-m) / n - m ] - [1 / n + m] - [ 1 / n - m]

    right?
     
  7. Aug 25, 2009 #6

    Dick

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    I would write that as cos(2*pi*(n+m))/(n+m) - cos(2*pi*(n-m))/(n-m) - [(1/(n+m)-1/(n-m)]. Use enough parentheses and brackets so your meaning is clear. But, yes, I think you mean the right thing. Now what's cos(k*2*pi) where k is an integer?
     
  8. Aug 25, 2009 #7
    the integral of cos(k*2*pi) is -sin(k*2*pi)/(2*pi)
     
  9. Aug 25, 2009 #8

    Dick

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    I'm not asking you what the integral is. I'm asking you what the value of cos(k*2*pi) is where k is an 'integer' - not 'integral'. If k=1, that's cos(2*pi), if k=2, that's cos(4*pi), if k=3 that's cos(6*pi) etc etc. I'm asking you this because (n+m) and (n-m) are integers.
     
  10. Aug 25, 2009 #9
    oh well in that case cos(k*2*pi) = 1
     
  11. Aug 25, 2009 #10

    Dick

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    Right, so what's the value of the integral now that you know what all the terms are?
     
  12. Aug 25, 2009 #11
    I think it's [(1/(n+m)-1/(n-m)] - [(1/(n+m)-1/(n-m)] which I think cancels out?
     
  13. Aug 25, 2009 #12

    Dick

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    Yes, it's zero. That means f(t)=sin(nt) and g(t)=cos(mt) are orthogonal over the interval [0,2*pi] since the integral of f(t)*g(t) over that interval is zero. There is a minor technical point. What happens if n=m or n=(-m)? Then your expression has a zero in the denominator. Can you see why that's not really a problem?
     
  14. Aug 25, 2009 #13
    yeah I think so, they still cancel out even though 1/0 tends to infinity
     
  15. Aug 25, 2009 #14

    Dick

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    That's dangerous reasoning. The point is that if e.g. n=m, then sin((n-m)*t)=sin(0*t)=0. The cos((n-m)t)/(n-m) term isn't in the antiderivative. If you are going to be formal about the proof the you might want to make those special cases.
     
  16. Aug 25, 2009 #15
    ahhh I see. Thanks you been really helpful
     
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