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Fourier Series

  1. Feb 18, 2010 #1
    1. The problem statement, all variables and given/known data
    f(x) = sin(x) for 0[tex]\leq[/tex]x<[tex]\pi[/tex]. Extend f(x) as an even function . Obtain a cosine Fourier series for f.


    2. Relevant equations
    [tex]a_{0}[/tex]/2 + [tex]\sum[/tex] [tex]a_{n}[/tex]cos(nx)


    3. The attempt at a solution
    So as far as I know, to extend sin(x) as an even function you have to make f(x)=-sin(x) for [tex]-\pi\leq[/tex]x<0 and then just use that to integrate for an but this gives a series without an a0 term which the question points to it having. What have I done wrong?
     
  2. jcsd
  3. Feb 18, 2010 #2
    Have you integrated correctly?

    Did you remember that the integral over [-pi,pi] breaks to [-pi,0] where it's -sin(x) and [0,pi] where it's sin(x) (or alternatively, the integral over [-pi,pi] is double the integral over [0,pi])

    Can you please show us your calculation?
     
  4. Feb 18, 2010 #3
    sure.
    As you've said, I've just used:
    [tex]\frac{2}{\pi}[/tex][tex]\int^{\pi}_{0}[/tex]sin(x) dx
    so integration gets
    [-cos(x)[tex]]^{\pi}_{0}[/tex].
    expanding gives [1 - 1] so I lose the [tex]a_{0}[/tex] term. Is that correct?

    Oops, i see what I did. Forgot the - - so it should actually be 4/pi. Is that correct?
     
  5. Feb 18, 2010 #4

    vela

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    Yeah, that's right.
     
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