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Fourier Derivation

  1. Mar 11, 2015 #1
    during these first few steps, where did the constants in front of the integrals come from for a_0,a_n, b_n?


    http://i.imgur.com/rky0mdf.png





    (wasn't sure whether to post this as a separate topic or back with the other one)
     
  2. jcsd
  3. Mar 11, 2015 #2

    mfb

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    From the integration range. Imagine a constant function, for example. The integral over the function from -lambda/2 to lambda/2 is then given by lambda*f(x). In order to match the definition above, you have to divide by lambda to get it right. The same applies to the sine and cosine terms, but in a more complicated way for the factor of 2.
     
  4. Mar 11, 2015 #3
    okay that makes sense. but for the a_n and b_n terms, multiplying by two; i've see this is my textbook, but they did that when they went from zero to lambda/2, and then multiplied by two, since they were dealing with an even function. but that doesn't seem to be what's happening here. could you elaborate on the more complicated way for obtaining the two factor?
     
  5. Mar 11, 2015 #4

    mfb

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    It comes from the fact that ##\int_0^{2\pi} sin^2 (t) dt = \frac{1}{2} 2 \pi##. This is different from the previous case where you had ##\int_0^{2\pi} 1^2 dt = 2 \pi##.
     
  6. Mar 12, 2015 #5
    am i on the right track for deriving the coefficient?


    http://i.imgur.com/RzeR8dL.jpg
     
  7. Mar 12, 2015 #6
    Try and calculate ##\int_{-\lambda/2}^{\lambda/2} f(x)\cos(k_n x) dx ##. Remember the properties of the cosine and sine basis.
     
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