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Aobut delta and Heaviside function...

by eljose
Tags: aobut, delta, function, heaviside
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eljose
#1
Jul15-05, 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]
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qbert
#2
Jul15-05, 03:11 PM
P: 185
H(ax) = H(x).
Crosson
#3
Jul15-05, 06:04 PM
P: 1,295
[tex] (H(ax))' = a H'(ax) = a \delta (ax) =\delta (x) = H'(x) [/tex]


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