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Aobut delta and Heaviside function... 
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#1
Jul1505, 01:54 PM

P: 501

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
Jul1505, 03:11 PM

P: 185

H(ax) = H(x).



#3
Jul1505, 06:04 PM

P: 1,295

[tex] (H(ax))' = a H'(ax) = a \delta (ax) =\delta (x) = H'(x) [/tex]



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