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Cosine series of sine

  1. Oct 16, 2005 #1
    I tried to find the cosine series of the function [tex]f(x) = \sin x[/tex], using the equation below:

    [tex]S(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx)[/tex]
    where: [tex]a_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \cos(nx) dx[/tex]

    I found:

    [tex]a_0 = \frac{4}{\pi}[/tex]
    [tex]a_n = \frac{2 }{\pi (1 - n^2)} (\cos(n \pi) + 1)[/tex]

    Therefore:

    [tex]S(x) = \frac{2}{\pi} (1 - 2 \sum_{n=1}^{\infty} \frac{\cos(2nx)}{4n^2 - 1})[/tex]

    Making the graph of the first terms of the function above on my calculator, I noticed that this is actually [tex]| \sin x |[/tex], rather than just [tex]\sin x[/tex]. Why does this happen? Is there a way of figuring out the sum not using a calculator?

    Thanks
     
    Last edited: Oct 16, 2005
  2. jcsd
  3. Oct 16, 2005 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    sinx is nonegative between 0 and pi, so sinx=|sinx| there.
     
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