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    • Thread Starter

    Hi, so far for an online problem I have correctly worked out the coefficients a0, an and bn of a Fourier series to be:
    a0 = 5*pi
    an = (1/pi)((2/(n^2)*(1-cos(n*pi)))+((12/(n^2))*(cos(n*pi)-1)))
    bn = ((-14/n)*cos(n*pi))

    Where the series if defined for (-pi to pi)

    F(x) = (2x for >= -pi to <0) and (12 for =>0 and < pi)

    The part of the question I am stuck on requires me to calculate Ff(pi) Which I understand is not defined from F(x) so I have then tried to use the formula (F(x+)+Fx-))/2 as the point is discontinuous, this gives me a value of 7pi which is apparently wrong.

    The Fourier series itself is defined as Ff(x) = 1/2*a0 + (sigma n=1 to infinity)an*cos(nx) + (sigma n=1 to infinity)bn*sin(nx)

    Sorry I know this may be difficult to understand but I don't know how to use any type of mathCad or editor or anything like that on forums. Thanks in advance, Kam.
    • Thread Starter

    Ok I have now obtained the right answer by a fluke; 5pi, but can anyone tell me why this is the answer as I do not understand where it has come from..
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Updated: February 13, 2010
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