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Fourier Series

  1. Jul 22, 2008 #1

    G01

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

    Hi everyone. I have a Fourier series problem from an old Physics GRE that I would like to discuss.

    Here is a link to the problem and a solution. The answer they get is correct, but their explanation is not:

    http://grephysics.net/ans/9277/39


    2. Relevant equations

    Fourier Analysis, but we should be able to solve this without too much math.

    3. The attempt at a solution

    Since, you only have 1min. 15s for every problem on the test, usually these problems can be solved in that length of time.

    So, I am trying to find a way to get to the correct answer by using reasoning and elimination instead of actually integrating to find the correct coefficients (time consuming). Here is where I am at:

    You can eliminate all choices except for A and B since the terms in the series must be odd functions since the function we are expanding is odd.

    Now, I can't seem to find a good way to narrow the answer down to B, which is the correct answer. Their explanation does not make sense, since if A is trivially 0 then so is B.

    Does anyone here have any ideas? I have been staring at this for quite some time. Yet, I can't find a good way to eliminate A, except by brute calculation of the series. Any thoughts and help will be appreciated. Thank you.
     
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  3. Jul 22, 2008 #2

    Dick

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    Well, it is sort of obvious that the integral of the function times sin(2wt) vanishes, right? That rules out A. The 'clue' is clearly bogus. And it took me considerably longer the 1min 15s for that, so I guess I lose.
     
  4. Jul 23, 2008 #3

    G01

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    Yes. That makes sense Dick. Thanks for the help. I really hate these problems. Obviously this does not test your ability with Fourier series, but instead tests your ability to find a work-around alternative to a real solution in a short time. Well, if the ETS actually had tests which were good indicators of physics ability, life would just be too easy.:rolleyes:
     
  5. Jul 23, 2008 #4

    Dick

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    It would have been much easier if I hadn't read the bogus solution first.
     
  6. Jul 23, 2008 #5

    G01

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    That website is hit and miss. They do have solutions to every released physics GRE question, but sometimes the solutions are not correct at all. That is one of the more blatant examples. Sometimes the website has mistakes which are very subtle and almost believable. You can see what happens to a "homework" website that does not have moderation like PF has.
     
  7. Jul 23, 2008 #6
    Ok, well I'd be guessing here. I did the "brute force method" to find Bn.
    Bn=4/npi.

    If I put this into Sum{Bn*sin(npi*t)}[1,2pi] I will get all zeros,ok we already know this so, that elimanates (A).

    Look at (A) then look at (B). Try playing with the indices of (B), if i=0,...inf 1/(2i+1)

    set m=1,...inf+1 1/(2m).

    Geez, what I am TRYING to say is,we all know that our choice of Bn*sin(npi*t) 4ALL 'n' yields a zero. Wait a minute!

    Just thought of something. Consider convergence, I mean |sin(x)|<=1,correct?
    so if that be the case,then why not use these??


    |sin(2n+1)|<=1 4ALL 'n' && so is |sin(n)|<=1.


    look at the sums (A) & (B). Which converges faster??

    1/(n) || 1/(2n+1)? For a 90 second problem,this might help.

    *Always think in simplest terms*
     
  8. Jul 23, 2008 #7
    Oh boy,what a blunder!

    (A) is ZERO

    Thus 0<1/(2i+1), so (B) would be by default, the correct answer. Bn=4/(2i+1)*pi

    So Sum{Bn*sin(2i+1)*wt}[0,inf] <=|sum{Bn}|[0,inf]
     
  9. Jul 23, 2008 #8
    But Jon, if (A) were zero,then that point is moot. Ok,ok,ok.

    |1/n| < |1/(2n+1)|, I state my case.
     
  10. Jul 23, 2008 #9

    Mute

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    There's a discussion on that page that gives the easy way to spot that it's B:

    [tex]V\left(t + \frac{\pi}{\omega}\right) = -V(t),[/tex]

    hence, the answer cannot be A because it contains even multiples of [itex]n\pi/\omega[/itex], and so [itex]V(t + \frac{\pi}{\omega}) \neq -V(t),[/itex] for choice A.
     
  11. Jul 24, 2008 #10
    so mute what you are saying is (A) is NOT half wave symmetrical?
    Therefore, by checking with the same rule,resolves the case for (B)?
    Thanks
     
  12. Jul 25, 2008 #11

    Mute

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    Yes. Choice A contains terms [itex]\sin(2n\omega t)[/itex].

    [tex]\sin\left(2n\omega \left(t + \frac{\pi}{\omega}\right)\right) =
    \sin\left(2n\omega t + 2n\pi\right) = \sin(2n\omega t)[/tex]

    Hence choice A does not have the [itex]V\left(t + \frac{\pi}{\omega}\right) = -V(t)[/itex] symmetry that the waveform shown in the problem has, whereas because choice B has only odd terms and so you get [itex]\sin\left(((2n+1)\omega t + (2n+1)\pi\right) = -\sin((2n+1)\omega t)[/itex] for every term in the sum, and so choice B does have the required symmetry.
     
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