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Finding the function given a Fourier Series.

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data
    From the Fourier series


    [tex]\frac{1}{2} - \frac{1}{4}cos(x) + \sum\frac{(-1)^{n}}{1-n^{2}}cos(nx)[/tex]


    of


    (1/2)x*sinx on [tex][-\pi,\pi][/tex], find the function whose Fourier Series on [tex][-\pi,\pi][/tex] is


    [tex]\frac{3}{4}sin(x) - \sum\frac{(-1)^{n}}{n-n^{3}}sin(nx)}[/tex]

    Both sums go from n=2 to n=infinity. The latex stack was showing up weird.
    2. Relevant equations


    3. The attempt at a solution
    Well, this seems like a fair bit of intuition. I tried to plug in F-Series formula for sin(x) and cos(x) into both F-Series but didn't get anywhere. I then realized that the second F-Series bears some resemblance to the integration of the first series. For example, the indefinite integral of the sum in the first F-Series is exactly the sum in the second series, aside from a constant. The integral of the -(1/4)cos(x) term yields (1/4)sinx, which is close and hence here has been everything I've tried.

    Any suggestions to get me going in the right direction?

    I also tried setting the a_n coefficient from the first F-Series equal to n*b_n of the second, and then I had the integrals equated but I can't really do much from there. Differentiating both integrals just gives 0= 0.

    EDIT: Sorry, I was missing "sin(x)" in my second F-Series Formula.
     
    Last edited: Oct 15, 2009
  2. jcsd
  3. Oct 15, 2009 #2
    So no one has any ideas?
    BTW this is a bump because I made a huge mistake on the latex for second F-Series.
     
  4. Oct 15, 2009 #3
    well it looks to me like you could consider the fourier series to the the resultant waveform whn a pure wave form is applied as the x in a f(x).
    Imagine that that f(x)=x^2. Then the pure waveform is modulated by f(x) so that where we had sin(x) we now have (sin(x)). Now we apply the appropriate trig identity:
    (sin(x))^2 = (sin(x))*(sin(x))= sin(a)*sin(B) where A=B (just so you recognise it!)
    sin(A)*sin(B)=(1/2)(cos(A+B)-cos(A-B))
    (sin(x))^2=(1/2)*(-cos(2*x)+cos(0)) (look out for that negative on the first cos!)
    The fourier series you quote is likely, I believe, to have resulted from such considerations at a higher power than 2. Such functions arise for instance when a small varying current passes through a silicon diode; the output voltage may well be represented by a function like this.
     
  5. Oct 15, 2009 #4

    Dick

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    Science Advisor
    Homework Helper

    You were already on the right track. If you call your infinite series with cos(nx) C(x) and your infinite series with sin(nx) S(x) then the first relation is (1/2)-cos(x)/4+C(x)=(1/2)x*sin(x). The unknown function is 3*sin(x)/4-S(x). As you've noticed S'(x)=C(x). Solve the first relation for C(x), substitute C(x)=(S'(x)) and integrate to find S(x). Now you only have a constant of integration to find.
     
    Last edited: Oct 15, 2009
  6. Oct 15, 2009 #5
    Hehe wow I feel pretty dumb, thanks for the help guys.
     
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