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Integral and Fourier serie

  1. Aug 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Let [tex]u(t)=2-\cos(t)+\sin(2t)- \cos(3t)+ \sin(4t)[/tex]

    Evaluate:

    [tex]\int_0^{2\pi}u^2(t)\mbox{d}t[/tex]

    2. Relevant equations



    3. The attempt at a solution

    Sorry, I don't have any idea :(... As I can see

    [tex]\int_0^{2\pi}u^2(t)\mbox{d}t[/tex]

    is similar to the first term of Fourier Serie of the function u^2(t)...
     
  2. jcsd
  3. Aug 23, 2013 #2
    You might consider using Parseval's theorem.
     
  4. Aug 23, 2013 #3

    HallsofIvy

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    I don't see any reason to think about "Fourier Series" at all- just integrate directly:
    [itex]\int_0^{2\pi} 4+ cos^2(t)+ sin^2(2t)+ cos^2(3t)+ sin^2(4t) dt[/itex]
    is easy.
    [itex]2\int_0^{2\pi}-2cos(t)+ 2sin(2t)- 2cos(3t)+ 2sin(4t)dt[/itex]
    [itex]2\int_0^{2\pi}- cos(t)sin(2t)- cos(t)cos(3t)+ cos(t)sin(4t)dt[/itex]
    [itex]2\int_0^{2\pi}-sin(2t)cos(3t)+ sin(2t)sin(4t)- cos(3t)sin(4t)dt[/itex]

    You may need the trig identities for cos(a)cos(b), sin(a)sin(b), and sin(a)cos(b).
     
  5. Aug 23, 2013 #4
    HallsOfIvy,

    I think you just squared each term, rather than squaring the whole expression?
     
  6. Aug 23, 2013 #5

    LCKurtz

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    ..along with the fact that ##u(t)## is its own Fourier Series.
     
  7. Aug 23, 2013 #6
    Ok, thanks for your hints!

    So

    [tex]\int_{0}^{2\pi}u^2(t)\mbox{d}t= 12\pi[/tex]

    right? :)
     
  8. Aug 23, 2013 #7

    Dick

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    Right. But if you think about orthogonality relations, you only have to evaluate the first integral Halls gave in post 3. The rest are obviously 0. And that one doesn't take much calculation either if you remember that the 'average value' of sin(nt)^2 or cos(nt)^2 where n>1 is 1/2.
     
    Last edited: Aug 23, 2013
  9. Aug 23, 2013 #8

    LCKurtz

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    Actually, you don't have to evaluate any integrals. The Fourier coefficients are sitting right there in the function u(t).
     
  10. Aug 24, 2013 #9

    Dick

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    Sure, if you use Parseval's. I wasn't sure which track the OP was following.
     
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