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Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

Compute its cosine Fourier series ?

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  • 2 months ago
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    omega = pi/3.

    a0 is simply the average value of the function on the interval -3 to 3; which is the same as the average value of e^x on [0,3].  The integral from x = 0 to 3 of e^x dx is

    e^3 - e^0 = 19.086; divide by 3 and you get 6.362.  

    a0 = 6.362.

    an = (1/3) * integral from x = -3 to +3 of

    e^x * cos(n*pi*x/3) dx.  [call this LINE ONE]

    This should be integrated by parts.

    Let u = cos(n*pi*x/3) and dv = e^x dx, so

    du = (n*pi/3) [-sin(n*pi*x/3)] dx and v = e^x.

    Now we have 

    uv - integral of v du

    = e^x*cos(n*pi*x/3) +

    + (n*pi/3) integral of e^x*sin(n*pi*x/3) dx.

    A second application of integration by parts is necessary:

    u2 = sin(n*pi*x/3) dx and dv2 = e^x dx, so

    du2 = (n*pi/3)*cos(n*pi*x/3) dx and v2 = e^x.

    Now we have

    e^x*cos(n*pi*x/3) + n*(pi/3)*e^x*sin(n*pi*x/3) -

    n^2*(pi^2/9)*integral of e^x*cos(n*pi*x/3) dx.

    Note that the whole expression on the last two lines is equal to the expression on LINE ONE above, so now you can combine the two like terms and obtain

    [1 + n^2*pi^2/9] * integral of e^x*cos(n*pi*x/3) dx

    = e^x*cos(n*pi*x/3) + n*(pi/3)*e^x*sin(n*pi*x/3).

    Which means that the integral is equal to

    [e^x*cos(n*pi*x/3) + n*(pi/3)*e^x*sin(n*pi*x/3)] /

    [1 + n^2*pi^2/9].

    Now we can plug in the limits (3 and -3) to evaluate "an" --

    you get

    2[e^3*cos(n*pi) - 1*cos(0)]/[1 + n^2*pi^2/9].

    The sine went away because sin(n*pi) is always 0.

    The use of 0 as the lower limit, and the "2" outside the first square bracket, come about because f(x) was created by reflecting the function e^x through the y axis.

    We need to multiply both the numerator and the denominator by 9, obtaining

    an = 18[e^3*cos(n*pi) - 1]/(9 + n^2*pi^2).

    Obviously cos(n*pi) is just 1 or -1, depending on whether n is even or odd, so we could write:

    an = 18[e^3*(-1)^n - 1] / (9 + n^2*pi^2).

    That's my final answer to your third question.

    I am encouraged by the resemblance of my "an" to the expression that appears in your fourth and final question, but my head is spinning now.  I'm 75, that's my excuse...

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