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Even and Odd functions - Fourier Series

  1. Dec 5, 2014 #1

    MMS

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    Hello everyone,

    I know that the integral of an odd function over a symmetric interval is 0, but there's something that's bothering my mind about it.
    Consider, for example, the following isosceles trapezoidal wave in the interval [0,L]:

    proxy.php?image=http%3A%2F%2Fs27.postimg.org%2Fl51m2w9v7%2Ftrap.png

    When expressed in Fourier series, the coefficient multiplied by the cosine (a_n) drops, but to be honest, I don't see why. By looking at the graph above, the only thing I can tell is that there is reflection symmetry of the function when looked at from L/2 which clearly doesn't indicate an odd function but rather an even one. Also, the interval is symmetric only when looked at from L/2. So, I can't really see how the theorem of odd functions in a symmetric interval is applied (Unless this isn't why the coefficient a_n is zero :p).

    I do, however, see by looking at the graph that the function has a sine-like structure so the Fourier series is most probably going to be consisted of sines (Not sure though if that's a way to look at it. Correct me if I'm wrong).
    But even when I calculated the coefficient a_n, it didn't drop. Checked my calculations a couple of times and I'm yet to find anything wrong (if needed, I'll upload it).

    I'd be really happy if someone could get this straight for me.

    Thank you!

    P.S: wasn't sure whether to upload here or to a more physics-related section since I'm talking about waves and strings, move it if needed.
     
  2. jcsd
  3. Dec 5, 2014 #2

    OldEngr63

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    How does your function continue to the left of the origin? Is it simply the reflection in the line x = 0?

    If your function as show is the entire period, then the average value, a0, is pretty clearly not zero, and the function has that even term.

    Recall that some functions are even, some are odd, and some are neither. All functions can be resolved into even parts and odd parts, however. Looks to me (without doing the calcs) that your test function is neither even nor odd, but rather consists of both parts.
     
  4. Dec 5, 2014 #3
    By simple looking at this graph, one can observe it has much more similarity with one semicycle of a sine wave than a cosine.
     
  5. Dec 5, 2014 #4

    mathman

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    It would help if you wrote exactly what the cos and sin angle variables and the domain of integration you used.
     
  6. Dec 6, 2014 #5

    MMS

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    it is indeed not odd after checking (plugging -x into the function of each interval [0,L/3], [L/3,2L\3], [2L/3,L]).

    So, you're claiming that the Fourier series will include both coefficients a_n and b_n?

    Yes. I mentioned that in my post. The question is, why does the coefficient multiplied by the cosine (a_n) in the Fourier series drops? It didn't drop by calculation and you can't apply the integration theorem of an odd function in a symmetrical interval.


    assuming the maximum height it reaches is a:

    Untitled.png
     
    Last edited: Dec 6, 2014
  7. Dec 6, 2014 #6

    OldEngr63

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    When I have Maple evaluate your integrals, I do not get zero for a0, an, or bn. This means that your function is neither even nor odd.
     
  8. Dec 6, 2014 #7

    MMS

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    Weird.
    Maybe I'm missing something out here. The problem is presented as followed:

    "Suppose we arouse a string fixed in both ends in the following way: [The figure I posted above]
    Find all the Fourier coefficients for the series representing the shape of the string at t(time)=0."

    And in their solution they state that a_n=0 with no explanation.

    What am I missing here?
     
  9. Dec 6, 2014 #8

    OldEngr63

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    I think you may be missing that the book can be wrong.
     
  10. Dec 6, 2014 #9

    mathman

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    From your definition for an and bn, the bn terms should be 0, not an. The function is even around L/2. cos is even, while sin is odd over the interval [0,L].

    On the other hand if the function goes from -L to L and is odd, the story is just the opposite, since you would have an odd function.
     
  11. Dec 7, 2014 #10

    MMS

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    Unfortunately, it isn't with this problem...




    I haven't done this before but is expanding a range from [0,L] to [-L,L] to eliminate some coefficients (in this case a_n since it would be an odd function around 0 in a symmetric interval) or is this process not legit?
     
  12. Dec 7, 2014 #11

    mathman

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    If the function is odd, all cos coefficients = 0. I think you should go back to the original question. Is the picture a half wave and you are asked for a series for the full wave?
     
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