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Sine series for f(x) = { 0 x<2

  1. Oct 29, 2003 #1
    why is

    the sine series for f(x) = { 0 x<2 on 0,3
    2 x>2

    L=3

    cn=2/3 int b=3 and a =0 f(x) sin( nxpi/ L)dx
    = 2/3 int. b=2 a=0 (0) sin( pi n x/ 3) dx + 2/3 int b=3 a =2 (2) sin n x pi/3 dx
    = 0 + 4/3{-3n pi cos n pi x/3} x=3 and x=2 = 4/n pi { cos 2 n pi/3 - cos n pi }

    f(x)=sigma n=1 4/n pi { cos 2n pi/3 -(-1)^n sin n pi x

    cos 2 pi/3=-1/2....

    f(x)=4/pi( 2/3 sin pix/3-3/4sin2 pi x/3 +2/3sin 3 pi x/3...

    how did they get this? very confused
     
    Last edited by a moderator: Feb 7, 2013
  2. jcsd
  3. Oct 29, 2003 #2
    my guess would be differnt N values
     
  4. Oct 30, 2003 #3
    ok never mind i get it now
     
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