Proof of the derivative of delta function

[rocky3321]

The problem is to prove that δ'(ax) = (1/a)*(1/a)*δ'(x), where a is a constant. I tried applying the scaling theorem with the formal definition of δ'(x) but I can not get the second (1/a) term. Does anyone have some insight on this problem? Thank you...

 

[rocky3321]

I was able to figure it out, so you do not have to reply to this thread.

 

[tomcue]

I can offer some insight on this problem. The scaling theorem states that if f(x) is a function and c is a constant, then δ(cx) = 1/|c| * δ(x). This theorem can be applied to the problem at hand by letting c = 1/a. This would give us δ(1/a * x) = 1/|1/a| * δ(x), which simplifies to a * δ(ax) = a * δ(x).

To obtain the additional (1/a) term, we can use the chain rule for derivatives. The chain rule states that if y = f(g(x)), then y' = f'(g(x)) * g'(x). In this case, let y = δ(ax) and f(x) = δ(x). Thus, y' = δ'(ax) * a.

Substituting this into our previous equation, we get a * δ'(ax) = a * δ(x). Dividing both sides by a, we get δ'(ax) = δ(x).

Therefore, we have shown that δ'(ax) = (1/a) * (1/a) * δ'(x), where a is a constant. This result can also be verified by plugging in values for x and a and observing that both sides of the equation are equal.

I hope this helps with your understanding of the problem. If you have any further questions or would like to discuss this further, please let me know. Thank you.