Distribution question

1. Aug 24, 2008

mhill

let be the distribution

$$D^{m} \delta (x-a) D^{k} \delta (x)$$

my questions are , what happens whenever x=a or x=a ??

is this identity correct

$$\delta (x-a) = e^{-a D} \delta (x)= \sum_{n=0}^{\infty}(-a)^{n} \frac{D^{n}}{n!}\delta (x)$$

2. Aug 24, 2008

jostpuur

I don't understand the first question, but this

is correct if you using such test functions that the Taylor series work for them. It can be checked with integration by parts. If the Taylor series don't converge for the test functions, then this distribution thing stops working as well.

3. Aug 24, 2008

Hurkyl

Staff Emeritus
This doesn't make sense to me. What meaning did you intend?

4. Aug 24, 2008

Redbelly98

Staff Emeritus
Probably meant to say x=a or x=0, since the expression contains the terms δ(x-a) and δ(x-0).

5. Aug 24, 2008

Hurkyl

Staff Emeritus
I meant that the idea of "plugging in" a value for x doesn't appear to make sense in this context.

6. Aug 24, 2008

mhill

the idea is

$$D^{m} \delta (x-a) D^{k} \delta (x)$$ however the value

$$D^{m} \delta (-a) D^{k} \delta (0)$$ and $$D^{m} \delta (0) D^{k} \delta (a)$$

is not defined since delta functions are just oo

another question , how would we define $$\int_{-\infty}^{\infty}dx D^{m} \delta (x-a) D^{k} \delta (x)$$

also , under suitable test function f , then

$$< f | \delta (x-a) > = \sum_{n=0}^{\infty} (-a)^{n} \frac{ < \delta | D^{n} f>}{n!}$$

although it would make no sense , i think we could say

$$(2\pi ) i^{m}D^{m}\delta (0) = \int_{-\infty}^{\infty}dx x^{m}$$ which is divergent... although in Cauchy's principal value the integral should be 0 for m Odd

Last edited: Aug 24, 2008