1. Limited time only! Sign up for a free 30min personal 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!

Sinc function limit question

  1. Aug 12, 2012 #1
    Consider [itex]sin(vx)/x[/itex] as [itex]v[/itex] approaches infinity. Now, this becomes a delta function and I have seen graphs that show this function as v increases.

    My question is the following: I cannot quite see why the function [itex]sin(vx)/x[/itex] becomes zero if x≠0. sin(vx) is bounded and oscillates rapidly between -1 and 1 as x is changed. But the envelope of the function is still 1/x so how come it goes to zero for all x≠0? Can someone prove this result?

    Thank you.
     
  2. jcsd
  3. Aug 12, 2012 #2
    I think you're talking about sin(1/x) not, (sinx)/x

    Edit: Actually re-reading your post, I realize I have no idea what you're asking.


    sinx/x is in no way related to the delta function, and does not oscillate between anything, it's just a normal sinx that diminishes as you approach infinity.
    Sin(1/x) however does oscillate in a divergent manner as you approach 0, the limit is quite strange and nothing is quite like it.
     
  4. Aug 12, 2012 #3
    No, it is sin(vx)/x. As v gets larger, it approximates a delta function better. Now, with v→∞, I cannot see why the function is zero if x is non zero.

    A similar definition is here http://en.wikipedia.org/wiki/Sinc_function where there is a section on how it relates to the dirac delta function.
     
    Last edited: Aug 12, 2012
  5. Aug 12, 2012 #4

    Mute

    User Avatar
    Homework Helper

    The proper way to show that

    [tex]\lim_{v \rightarrow \infty} \frac{\sin(vx)}{x} = \delta(x)[/tex]

    is not to try and show that the left hand side is infinite when x = 0 and zero for non-zero x. That might be true of some representations of the delta function, but really what you need to show is that

    [tex]\lim_{v\rightarrow \infty} \int_{-\infty}^\infty dx~\frac{\sin(vx)}{x} f(x) = f(0).[/tex]

    Similarly you can show that if the integration region does not contain the origin, the integral is zero.

    That is the appropriate sense in which one should interpret "[itex]\lim_{v \rightarrow \infty} \frac{\sin(vx)}{x} = \delta(x)[/itex]".
     
  6. Aug 13, 2012 #5
    Thank you Mute. Didn't know that before but yeah, I can prove it according to your definition so I guess it's all good.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Sinc function limit question
  1. Question on limits (Replies: 16)

  2. Limit of a function (Replies: 5)

Loading...