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Fourier Series for (sin(x))^2

  1. Sep 9, 2013 #1
    Hey there!
    I'm trying to calculate the Fourier Series for sin2x on [-π, π]

    For a0 I found 1/2. (By determining the average value of the function on the interval)

    Since sin2x is even, I know that bn = 0.
    Now, for an.. The following link shows the integral I used to try to evaluate an.
    http://www.wolframalpha.com/...

    Since an and bn are both coming out to be zero, doesn't this imply that sin2x = a0 = 1/2? Will someone please show me where I am going wrong? Can't find anything online about this. Everyone just always uses the sin2 identity and calls it good.
     
    Last edited: Sep 9, 2013
  2. jcsd
  3. Sep 9, 2013 #2

    lurflurf

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    hint: Recall that
    (sin(x))^2=(1/2) (1-cos(2 x))

    as for the integral we have

    $$a_2=\frac{1}{\pi} \int _{- \pi}^\pi \! \sin^2(x) \cos(2x) \, \mathrm{dx}=\frac{1}{2\pi} \int_{-\pi}^\pi \! (1-\cos(2 x)) \cos(2x) \, \mathrm{dx}$$
     
  4. Sep 9, 2013 #3
    What can I say.... My mind is blown...

    When you take the integral of sin2(x)cos(nx) you get -[itex]\frac{4sin(πn)}{n^3-4n}[/itex]

    It is easy to assume that this is equal to zero for all values of n, but in hindsight it is clear that this does not apply for n = 2. :S!

    I'm utterly confused now. Just when I thought I was beginning to understand how to compute Fourier Series......
     
  5. Sep 9, 2013 #4

    mathman

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    For n = 2, use L'Hopital's rule. I got -2π.

    One could also go to basics, using cos2x = cos2x-sin2x
    Therefore sin2x = 1/2(1 - cos2x)
     
    Last edited: Sep 9, 2013
  6. Sep 12, 2013 #5

    vanhees71

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    It's clear that [itex]a_0=1[/itex] and [itex]a_1=-1/2[/itex], because as said in the previous posting from the double-angle theorem you get
    [tex]\sin^2 x=\frac{1}{2}[1-\cos(2 x)].[/tex]
    For the Fourier coefficients you indeed have
    [tex]a_n=\frac{1}{\pi} \int_{-\pi}^{\pi} \mathrm{d} x \sin^2 x \cos(n x)=\frac{4}{\pi(4n-n^3)} \sin (n \pi).[/tex]
    This is indeed [itex]0[/itex] for all [itex]n \in \mathbb{N} \setminus \{2 \}[/itex], and for [itex]n=2[/itex] you get by direct evaluation of the integral [itex]a_2=-1/2[/itex].

    Of course you can also take the limit [itex]n \rightarrow 2[/itex] by using de L'Hospital's rule, because it's a limit of the indefinite form 0/0:
    [tex]\lim_{n \rightarrow 2} a_n=\lim_{n \rightarrow 2} \frac{4}{\pi} \frac{n \pi \cos(n \pi)}{4-3n^2}=-\frac{1}{2}.[/tex]
     
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