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How do you prove this?

  1. Mar 11, 2005 #1
    Let [tex]\theta(x)[/tex] be a Step function:

    [tex]\theta(x) = 1[/tex] if x>0
    [tex]\theta(x) = 0[/tex] if x=<0

    Show that [tex]\frac{d\theta}{dx}=\delta(x)[/tex]

    [tex]\delta(x)[/tex] is a Dirac Delta function.
     
  2. jcsd
  3. Mar 11, 2005 #2

    arildno

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    Let f(x) be a comparison function so that [tex]\lim_{x\to\pm\infty}f=0[/tex]
    Furthermore, we have:
    [tex]\int_{-\infty}^{\infty}\theta(x)\frac{df}{dx}dx=\int_{0}^{\infty}\theta(x)\frac{df}{dx}dx+\int_{-\infty}^{0}\theta(x)\frac{df}{dx}dx=-f(0)[/tex]
    However, by using integration by parts and the infinity conditions on f, we have:
    [tex]\int_{-\infty}^{\infty}\theta(x)\frac{df}{dx}dx=-\int_ {-\infty}^{\infty}\frac{d\theta}{dx}fdx[/tex]
    That is, for arbitrary f, we have:
    [tex]f(0)=\int_{-\infty}^{\infty}\frac{d\theta}{dx}fdx[/tex]
    which suggests the introduction of the delta function.
    (Of course, this "proof" is as unrigourous as it can be..)
     
    Last edited: Mar 11, 2005
  4. Mar 11, 2005 #3
    Thanks a million!
     
  5. Mar 12, 2005 #4
    What do you mean-->is there a more rigourous proof?
     
  6. Mar 12, 2005 #5

    arildno

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    To focus on one aspect, Dirac's delta function isn't a proper function.
    Its very definition is, from a strict, mathematical perspective, meaningless.
    In order to make a mathematically sensible&rigourous definition of the Dirac "function", mathematicians have developed the tools of "generalized functions"&"distributions.
     
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