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Dirac Delta Function Proof 
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#1
Mar1512, 07:39 PM

P: 400

1. The problem statement, all variables and given/known data
See http://mathworld.wolfram.com/DeltaFunction.html I want to show (6) on that page. I can show it using (7), but we aren't supposed to do that. I already proved (5), and my prof says to use the fact that (5) is true to get the answer. 2. Relevant equations 3. The attempt at a solution Here's what I tried: δ(x^{2}  a^{2}) = δ((xa)(x+a)) I'm not sure how to use (5), because here a is not multiplying x. I'm not sure where to go from here. 


#2
Mar1512, 08:47 PM

P: 312

Imagine what the delta look like in the neighborhood of a and a, i.e., when one factor goes to zero, the other factor is pretty much constant over that entire neighborhood.



#3
Mar1512, 09:37 PM

P: 400

I don't really know what it looks like. I know that δ(x) is zero everywhere except at x = 0. At x = 0, it's infinity.
I know that δ(xa) is the same as above except that now it's infinity at x = a. But I don't know what δ(x^{2}) looks like. 


#4
Mar1512, 10:19 PM

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Dirac Delta Function Proof



#5
Mar1512, 10:31 PM

P: 400

I don't understand why it looks like that though. I am having problems visualizing it.
I don't get how you know what it looks like unless it's just δ(x) or δ(xa) by itself. 


#6
Mar1512, 10:37 PM

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#7
Mar1512, 10:55 PM

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Ohh okay, I see that near x = a, (x+a) is about 2a. So we're just making an approximation and plugging it into the delta function, is that right?
I wasn't sure what the delta function itself looked like, not what x+a looks like. 


#8
Mar1512, 11:03 PM

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#9
Mar1512, 11:19 PM

P: 400

Alright, that makes a lot more sense now. So basically, we're saying:
δ((xa)(x+a)) = δ((xa)*2a) + δ((x+a)*(2a)) Is it okay to do that because it's zero elsewhere (within the delta function)? 


#10
Mar1512, 11:24 PM

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#11
Mar1612, 12:50 PM

P: 312

The idea is δ(f(x)) is zero except at f(x0)=0, so all that matters is the local behavior of f(x) near x0, so you can approximate f(x) around x0 by f(x)≈f'(x0)(xx0). Since all the zeros of f(x) must be accounted for, you easily derive the general formula (7) mentioned in your original post. This is certainly not a formal proof, as Dick pointed out, but I think you can have a formal but still not rigorous proof by using a test function, i.e., try evaluate ∫ δ(f(x))*g(x) dx and see what you get.



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