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Fourier Series

  1. Apr 13, 2010 #1
    I once wanted to find the nth order Fourier approximation for [tex]f(x)=x[/tex]. Since this function is odd, the projections on all cosines will be zero, hence it will be expressed through the sines only. So I just needed to find the sine coefficients.

    The problem is that I checked the answer to this problem in two different textbooks, each one giving a different answer from the other.

    • The first book calculates the coefficients like this:
    [tex]b_k= \frac{1}{\pi} \int^{2 \pi}_{0} f(x)sin(kx)= \frac{1}{\pi} \int^{2 \pi}_0 xsin(kx) dx =- \frac{2}{k}[/tex]

    • The second book calculates the coefficients this way:
    [tex]b_k = (f(x),sin (k \pi x))= \int^1_{-1} x sin (k \pi x) dx= -x\frac{cos (k \pi x)}{k \pi} \right]^1_{-1} + \int^1_{-1} \frac{cos (k \pi x)}{k \pi}dx [/tex][tex]= -2 \frac{cos k \pi}{k \pi}=-(-1)^k \frac{2}{k \pi}[/tex]

    So which one is correct? I mean, they both are probably correct but I don't understand why each of these books use different methods and end up with different answers. Any explanation is very much appreciated.
  2. jcsd
  3. Apr 13, 2010 #2


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    Those two expressions are for the series of different functions. One is the periodic extension of f(x) = x on (-1,1) and the other the periodic extension of f(x) = x on (-π,π). One is a sawtooth of period 2π and the other a sawtooth of period 2. They happen to agree on (-1,1).
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