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Dirac function

  1. Mar 14, 2010 #1
    Dirac function :(

    Hello everyone...

    I have some triple with my PDEs course especially with the Dirac function.
    How can I prove it is discontinuous function?
    I do not know where can I start...
    Could somebody help me, please.
     
  2. jcsd
  3. Mar 15, 2010 #2
    Re: Dirac function :(

    Well, for a start, the Dirac "function" is not even considered a function. It doesn't qualify for it. There are two approaches two define it. You can considere it as a limit of several continuous sequences of functions, for example:

    [tex]\delta_{n}=\frac{1}{n\sqrt{\pi}} e^{-x^{2}/n^{2}} , n\rightarrow 0 [/tex]

    is one such sequence. The limit, however, is not defined as a function. The right name is distribution. See http://en.wikipedia.org/wiki/Distribution_(mathematics)
     
  4. Mar 15, 2010 #3

    Redbelly98

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    Re: Dirac function :(

    (Moderator's note: thread moved from "Differential Equations")

    Start with the definition of a continuous function. At least one of the conditions is not satisfied, at at least one value of x, for a function to be discontinuous.
     
  5. Mar 15, 2010 #4
    Re: Dirac function :(

    It's pretty simple really. Just prove that the left and right limits at 0 are 0 (pretty easy), so in order to be continuous, f(0) must be 0, but it isn't, QED.
     
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