Aobut delta and Heaviside function

  • Thread starter eljose
  • Start date
  • #1
eljose
492
0
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]
 

Answers and Replies

  • #2
qbert
185
5
H(ax) = H(x).
 
  • #3
Crosson
1,259
4
[tex] (H(ax))' = a H'(ax) = a \delta (ax) =\delta (x) = H'(x) [/tex]
 

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