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Aobut delta and Heaviside function

  1. Jul 15, 2005 #1
    we know that [tex]\delta(xa)=(1/a)\delta(x) [/tex] if the dirac,s delta satisfies this then given the function H(ax) with H the Heaviside step function what relationship is there between H(ax) and H(x) with

    [tex] \frac{dH}{dx}=\delta(x) [/tex]
     
  2. jcsd
  3. Jul 15, 2005 #2
    H(ax) = H(x).
     
  4. Jul 15, 2005 #3
    [tex] (H(ax))' = a H'(ax) = a \delta (ax) =\delta (x) = H'(x) [/tex]
     
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