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Dirac Delta Function - Fourier Series

  1. Nov 23, 2015 #1

    1. The problem statement, all variables and given/known data

    Find the fourier series of

    ##f(x) = \delta (x) - \delta (x - \frac{1}{2})## , ## - \frac{1}{4} < x < \frac{3}{4}##
    periodic outside.



    2. Relevant equations

    ##\int dx \delta (x) f(x) = f(0)##

    ##\int dx \delta (x - x_0) f(x) = f(x_0)##


    3. The attempt at a solution

    I am struggling to visualise this function, which is making it hard to tell if my series makes sense. Plus I find the delta function a little confusing to reason about.

    The function has a period of 1.

    ##a_0 = 2 \int^{3/4}_{-1/4} \delta (x) dx - 2 \int^{3/4}_{-1/4} \delta (x - \frac{1}{2}) dx = 0##

    ##a_n = 2 \int^{3/4}_{-1/4} \delta (x) \cos (n \pi x) dx - 2 \int^{3/4}_{-1/4} \delta (x - \frac{1}{2}) \cos (n \pi x)dx = 2 - 2 \cos (\frac{n \pi}{2})##

    ##b_n = 2 \int^{3/4}_{-1/4} \delta (x) \sin (n \pi x) dx - 2 \int^{3/4}_{-1/4} \delta (x - \frac{1}{2}) \sin (n \pi x) dx = -2 \sin (\frac{n \pi}{2})##

    I think those are my fourier coefficients, and I couldn't find a nice way to express them, so I think they're ok left as trig functions.

    Does it look ok?

     
  2. jcsd
  3. Nov 23, 2015 #2

    HallsofIvy

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    [itex]cos(\frac{n\pi}{2})[/itex] is 0 if n is odd, 1 if n is an even multiple of 2 and -1 if n is an odd multiple of 2. Sin(x) is 0 if n is even, 1 if n is of the form 4k+1 for some k and -1 if it is of the form 4k-1 for some k. So [tex]\sum_{n= 0}^\infty(2- 2cos(\frac{n\pi}{2})cos(n\pi x)+ 2 sin(\frac{n\pi}{2})sin(n\pi x)= \sum_{n=0}^\infty [(2- 2(-1)^{n})cos(2n\pi x)+ 2(-1)^n sin(((2n+1)/2]\pi x)][/tex]
     
  4. Nov 23, 2015 #3
    Thanks a lot.

    That's incredibly helpful and clear.
     
  5. Nov 24, 2015 #4

    vela

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    You can also use some trig identities to simplify
    $$a_n \cos n\pi x + b_n \sin n\pi x = 2\cos (n\pi x) - 2 \left[\cos \left( \frac{n \pi}{2}\right) \cos (n\pi x) + \sin \left(\frac{n \pi}{2}\right) \sin (n\pi x) \right]$$ to get
    $$-4 \sin \frac{n \pi}{4} \sin\left[n\pi \left(x - \frac 14\right) \right]$$
     
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