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    • Thread Starter

    My question says to find the fourier series of e^|x| between -1<x<1 so period = 2, therefore L = 1
    I know its an even function, so the sin part coefficient = 0
    To find the cos coefficient, I calculate:

    2/L *integral(dx sin(nPIx/L) f(x)

    In their solution they have given f(x) as just e^-x, but isnt that only considering x>0? What about the x<0 section of the graph? I was thinking use two cases.

    Since f(x) and cos(x) are both even functions, the x<0 section is a mirror image of the x>0 side, so the integral over [-1,1] is just twice that over [0,1]. (This is where the 2 comes from in what you've posted).
    • Thread Starter

    oh yeah that makes sense, thanks
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Updated: April 11, 2011

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