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Math Help - Fourier Series

  1. #1
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    Fourier Series

    I want to find the nth order Fourier approximation for f(x)=x. 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 need to find the sine coefficients.

    The problem is that I checked the answer to this problem in two different textbooks, the solutions in the books are totally different from each other.

    • The first book calculates the coefficients like this:
    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}

    • The second book calculates the coefficients this way:
    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} ]^1_{-1} + \int^1_{-1} \frac{cos (k \pi x)}{k \pi}dx = -2 \frac{cos k \pi}{k \pi}=-(-1)^k \frac{2}{k \pi}

    So which one is correct? I mean, they both are probably correct but I don't understand why they use different methods and end up with different answers. Any explanation is very appreciated.
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  2. #2
    MHF Contributor

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    They are using two different "normalizations" because they are calculating the Fourier series over two different intervals.

    Remember that you cannot actually find a Fourier series for "f(x)= x". You can only find Fourier series for periodic functions.

    In the first case they are finding the Fourier series for the function that is equal to x for 0\le x\le 2\pi and then repeats that value with period 2\pi.

    In the second case they are finding the Fourier series for the function that is equal to x for -1\le x\le 1 and then repeats that value with period 2.
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