Dirac Delta Function Proof

by jumbogala
Tags: delta, dirac, function, proof
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 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(x2 - a2) = δ((x-a)(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.
 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.
 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(x-a) is the same as above except that now it's infinity at x = a. But I don't know what δ(x2) looks like.
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Dirac Delta Function Proof

 Quote by jumbogala 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(x-a) is the same as above except that now it's infinity at x = a. But I don't know what δ(x2) looks like.
Near x=a, δ((x-a)(x+a)) pretty much looks like δ((x-a)*2a). That's sunjin09's point.
 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(x-a) by itself.
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 Quote by jumbogala 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(x-a) by itself.
Near x=a, (x+a) is nearly 2a. You can't visualize that?
 P: 400 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.
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 Quote by jumbogala 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.
Yes, I think you are ok with hand waving through this. Near x=(-a) the value of (x-a) is nearly -2a. So split it into two delta functions at the two values where x^2-a^2 vanishes.
 P: 400 Alright, that makes a lot more sense now. So basically, we're saying: δ((x-a)(x+a)) = δ((x-a)*2a) + δ((x+a)*(-2a)) Is it okay to do that because it's zero elsewhere (within the delta function)?
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 Quote by jumbogala Alright, that makes a lot more sense now. So basically, we're saying: δ((x-a)(x+a)) = δ((x-a)*2a) + δ((x+a)*(-2a)) Is it okay to do that because it's zero elsewhere? It seems a little odd to split a multiplication up like that.
Yes, I think it's ok to do that because it's zero elsewhere. It's not a formal proof, but the answer is correct.
 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)(x-x0). 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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