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Fourier Cosine Series

  1. May 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Screen_shot_2012_05_18_at_5_53_44_PM.png

    3. The attempt at a solution
    So after I have properly extended the series, I want to find the A_n coefficients. In the case of A_0 I get 1/2 -- good, great. That's what the book gets.

    Now for the general A_n:

    [itex]A_n = \frac{1}{2} \int^2_{-2} (1)cos(\frac{n\pi x}{2}) dx [/itex]
    because, in the piecewise function, the portions from -2 to -1 and from 1 to 2 are zero, we can just write this as

    [itex]A_n = \frac{1}{2} \int^1_{-1} (1)cos(\frac{n\pi x}{2}) dx [/itex]
    [itex]= \frac{1}{2}(\frac{2}{n\pi} sin(\frac{n\pi x}{2}) |^1_{-1} = \frac{2}{n\pi} sin(\frac{n\pi}{2}) n ≥ 1[/itex]

    and because integers n's will just cause the sin term to alternate between -1 and 1:

    [itex] = \frac{(2)(-1)^{n+1}}{n\pi} n≥1[/itex], and this is my general term for the coefficients. But this isn't what the book gets, so what have I done wrong?
     
  2. jcsd
  3. May 18, 2012 #2

    Ray Vickson

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    If the period is 4, we need to know the values of f(x) between x = 2 and x = 4 (or, perhaps, between x = -2 and x = 0). What are those values?

    RGV
     
  4. May 18, 2012 #3
    Since it says to use a cosine series, I presume that the question intends for us to use an "even" function extension for the piecewise. Thus, the values for f(x) = 0 between -2<x<-1 and f(x) = 1 for -1 < x < 0. Assigning f(x)'s extension in such a way makes the function even and thus permits us to find a cosine series.
     
  5. May 18, 2012 #4

    Ray Vickson

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    OK, that seems fair. You say your answer differs from that in the book. What is the book's answer? Is it, perhaps, just a matter of some different normalization convention? I don't see anything obviously wrong with your computation.

    RGV
     
  6. May 18, 2012 #5
    This is the books answer:

    Screen_shot_2012_05_18_at_8_29_38_PM.png

    my answer will be [itex]f(x) = \frac{1}{2} + \sum_{n=1}^∞ ( \frac{(2)(-1)^{n+1}}{n\pi}) cos(\frac{n\pi x}{2})[/itex]

    But they seem to think that we're skipping some terms (hence the 2n - 1 things), but I don't have any terms zeroing so I don't need to make that adjustment.
     
  7. May 18, 2012 #6
    So I just tried to plot the first couple of terms in the series in MATLAB. I've attached the resulting graph in this post. As you can see, my answer isn't doing the right thing -- it's not correctly approximating the solution. Now, granted, this is only seven terms, but it doesn't seem to be doing very well. So I'm thinking that I must have a mistake somewhere in this work...

    Screen_shot_2012_05_18_at_9_31_34_PM.png
     
  8. May 19, 2012 #7

    Ray Vickson

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    For even n ≥ 2 we will have [itex] \int_{-1}^1 \cos(n \pi x /2)\, dx = 0[/itex] because in each interval (-1,0) and (0,1) the positive and negative areas cancel. For example, when n = 2 the areas from x = 0 to x = 1/2 and x = 1/2 to x = 1 are opposite.

    RGV
     
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