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Generalized functions (distributions) problem - Mathematical physics

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Find a distribution [tex]g_n[/tex] which satisfies
    [tex] g'_n(x) = \delta(x - n) - \delta(x + n) [/tex]
    and use it to prove
    [tex] \lim_{n \to \infty} \frac{\sin{nx}}{\pi x} = \delta(x) [/tex]

    2. Relevant equations
    Nothing relevant comes up at the moment.

    3. The attempt at a solution
    Well the first part is pretty easy I think. The distribution would be
    [tex] g_n(x) = \theta(x - n) - \theta(x + n) = \left\{ \begin{array}{l l}
    -1 & \quad |x| < n \\
    0 & \quad |x| \geq n \\
    \end{array} \right.[/tex]

    The limit will indeed resemble a delta function when [tex]n[/tex] goes to infinity and π is probably just a normalization constant. But applying the two Heaviside functions to solve this has got me stumped.

    P.S. Gotta catch some sleep, I will be back in 7 hours hopefully with some ideas to solve this.

  2. jcsd
  3. Feb 6, 2012 #2


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    g' looks like the Fourier transform of sine.
  4. Feb 6, 2012 #3
    Thanks Vela! That did the trick, I'll post the full solution later. :)
  5. Feb 6, 2012 #4
    Right guys so here it is, me and my buddies came up with this solution to the problem.

    So we have established that our distribution should be
    [tex] g_n(x) = \theta(x - n) - \theta(x + n) [/tex]
    Like Vela said, the Fourier transform of this sine resembles our distribution differentiated. So that means we have (using Euler equation for sine)
    [tex] \frac{d}{d\xi} \mathcal{F} \left\{ \frac{\sin{(nx)}}{\pi x} \right\}(\xi) = \frac{d}{d \xi} \int \frac{1}{\pi x} e^{i \xi x} \left( \frac{e^{inx} - e^{-inx} }{2i} \right) dx = \int \frac{i x}{2 i \pi x} e^{i\xi x} (e^{inx} - e^{-inx}) dx [/tex]
    [tex] = \frac{1}{2\pi} \int e^{ix(\xi +n)} - e^{ix(\xi - n)} dx = \frac{2\pi}{2\pi} \left( \delta(\xi + n) - \delta(\xi - n) \right) = \frac{d}{d\xi} \left( \theta(\xi + n) - \theta(\xi - n) \right) = \frac{d}{d\xi} (- g_n(\xi)) [/tex]
    If we look at the limit of our distribution. You can easily see that it will become unity. (You can plot those heaviside functions and see for yourself).
    [tex]\lim_{n \to \infty} (- g_n(\xi)) = 1 [/tex]
    Using inverse Fourier-Transform we have
    [tex] \mathcal{F}^{-1} \mathcal{F} \left\{ \frac{\sin{(nx)}}{\pi x} \right\}(\xi) = \mathcal{F}^{-1} \{1\}(x) = \delta(x) [/tex]
    Which ultimately proves what was supposed to prove
    [tex]\lim_{n \to \infty} \frac{\sin{(n x)}}{\pi x} = \delta(x) [/tex]

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