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Fourier Series

  1. Nov 27, 2009 #1
    I need to find the fourier series for the function f;

    0 if -[tex]\pi[/tex] [tex]\prec[/tex] x [tex]\leq[/tex] -[tex]\frac{\pi}{2}[/tex]

    1+x if -[tex]\frac{\pi}{2}[/tex] [tex]\prec[/tex] x [tex]\prec[/tex] [tex]\frac{\pi}{2}[/tex]

    0 if [tex]\frac{\pi}{2}[/tex] [tex]\leq[/tex] x [tex]\leq[/tex] [tex]\pi[/tex]

    I've never done a fourier series computation before so I don't really know if any of what I'm doing is correct.

    I've got than a(0)=1, a(n)=[tex]\frac{2sin\frac{n\pi}{2}}{n\pi}[/tex] and
    b(n)=[tex]\frac{2sin\frac{n\pi}{2}}{n^{2}\pi}[/tex] - [tex]\frac{cos\frac{n\pi}{2}}{n}[/tex]

    I know the formula for a fourier series but none of the examples I've seen are in the form of the a(n) and b(n) that I've got so I don't know where to go next.
    Could anyone help please? Thankyou.
  2. jcsd
  3. Nov 27, 2009 #2


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    You are going to use the fact that [itex]sin(n\pi/2)= 0[/itex] if n is even, aren't you? And that [itex]sin(n\pi/2)= 1[/itex] if n is "congruent to 1 mod 4" (i.e. n= 4k+ 1 for some integer k) while [itex]sin(n\pi/2)= -1[/itex] if n is "contruent to 3 mod 4" (i.e. n= 4k+ 3 for some k. That is, [itex]a_2= 0[/itex], [itex]a_3= -1/3\pi[/itex], [itex]a_4= 0[/itex], [itex]a_5= 1/5\pi[/itex], etc.

    Like wise [itex]cos(n\pi/2)= 0[/itex] if n is odd, [itex]cos(n\pi/2)= 1[/itex] if n is "congruent to 2 mod 4" (divisible by 2 but not by 4), and [itex]cos(n\pi/2)= -1[/itex] if n is "congruent to 0 mod 4" (divisible by 4). Notice that for all n, one of the [itex](2sin(n\pi)/2)/n\pi[/itex] or [itex]cos(n\pi/2)/n\pi[/itex] is 0 so you are not actually "subtracting".
  4. Nov 28, 2009 #3
    I can see the sequences that a(n) and b(n) make, but I can't write down a formula for them. I'm using trial and error really but I can't find any functions dependent on n that will describe the sequences a(n) and b(n) exactly. Is there a formula for caculating these type of functions?
  5. Nov 28, 2009 #4


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    You said that a(n)= [tex]\frac{2sin\frac{n\pi}{2}}{n\pi}[/tex]

    Since [itex]sin(n\pi/2)= 0[/itex] for n even, for a(n) to be non-zero, n must be of the form 2k+1 for some integer, k. But then [itex]sin(n\pi/2)= sin((2k+1)\pi/2)= sin(k\pi+ \pi/2)= 1[/itex] if k is even and -1 if k is odd. That is, [itex]sin((2k+1)\pi/2)= (-1)^k[/itex].

    So a(n)= 0 if n is even and, if n is odd, n= 2k+1, [itex]a(n)= a(2k+1)= 2(-1)^k/(2k+1)\pi[/itex]. You could write the Fourier cosine series as
    [tex]1+ \sum_{k=0}^\infty \frac{2(-1)^k}{(2k+1}\pi} cos((2k+1)x)[/itex].

    The sine series, [itex]\sum b_n sin(nx)[/itex] is a little more complicated but the same idea.
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