1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    Cheers
     
  2. jcsd
  3. Feb 6, 2012 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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]


    gg
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Generalized functions (distributions) problem - Mathematical physics
Loading...