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    f(x)=x, for 0≤x≤π

    The function is then defined on the interval −π≤x≤π by extending it as an even function. It is then defined on the whole real line by extending it as an 2π-periodic function.

    Calculate the Fourier series of f(x).

    How do I do this? I'm unsure how to tackle this as I don't know what interval I'm supposed to calculate the Fourier series for.
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    (Original post by victoriarty)
    f(x)=x, for 0≤x≤π

    The function is then defined on the interval −π≤x≤π by extending it as an even function. It is then defined on the whole real line by extending it as an 2π-periodic function.

    Calculate the Fourier series of f(x).

    How do I do this? I'm unsure how to tackle this as I don't know what interval I'm supposed to calculate the Fourier series for.
    I don't think it matters whether you choose −π≤x≤π or 0≤x≤2π. The coefficients should come out the same.
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    (Original post by victoriarty)
    f(x)=x, for 0≤x≤π

    The function is then defined on the interval −π≤x≤π by extending it as an even function. It is then defined on the whole real line by extending it as an 2π-periodic function.

    Calculate the Fourier series of f(x).

    How do I do this? I'm unsure how to tackle this as I don't know what interval I'm supposed to calculate the Fourier series for.
    Preferably you choose your integration region \int_{\alpha}^{\alpha + p} to be symmetric over x=0.

    Here, this is an even function so it will have a first term \displaystyle a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) .dx. Note that f(x) is an even function, therefore \displaystyle \int_{-\pi}^{\pi} f(x) .dx = 2 \int_0^{\pi} f(x) .dx which simplifies the integral down.

    Similarly, we have \displaystyle a_n = \frac{2}{\pi} \int_0^{\pi} f(x) \cos(nx).dx
 
 
 
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