Starting with the definition of the Dirac delta function,

In summary, the homework statement asks you to show that \delta( \sqrt{x}) equals 0. You start by defining the Dirac delta function and showing that \delta( \sqrt{x}) is equal to 0 if x equals x'. Next, you use integration to show that \delta( \sqrt{x}) equals 0 if x equals x-a. Finally, you use substitution to show that \delta( \sqrt{x}) equals 0 if x equals \sqrt{
  • #1
skrtic
15
0

Homework Statement



Starting with the definition of the Dirac delta function, show that [itex] \delta( \sqrt{x}) [/itex]um... i have looked in my book and looked online for a problem like this and i really have no clue where to start. the only time i have used the dirac delta function is in an integral with another function and never with it in this form, only like delta(x-a).looking for guidance.
 
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  • #2
skrtic said:

Homework Statement



Starting with the definition of the Dirac delta function, show that [itex] \delta( \sqrt{x}) [/itex]

Show that [itex]\delta( \sqrt{x}) [/itex] is what?
 
  • #3
sorry about that.

equal to 0
 
  • #4
How is the Dirac delta function defined? The problem asks you to start with this definition.
 
  • #5
from what i get out of the text. it is what replaces an inner product that vanishes if x doesn't equal x'.

I also know that it's integral is unity.
 
  • #6
[tex]\int_{-\infty}^{+\infty}dx\,f(x)\delta(x)=f(0)[/tex]

Now, if only the argument of the delta function was [itex]\sqrt{z}[/itex] instead of [itex]x[/itex] ... hmmm, how could we make this happen?
 
  • #7
let x= [itex] \sqrt{z} [/itex] then dx=(1/2)z^(-1/2)dz

and we get [itex] \int_{-\infty}^{\infty}dz z^{-1/2} f(z^{-1/2})\delta(z^{-1/2}) [/itex]
 
  • #8
is this a valid assumption

treat [itex] \delta(\sqrt{x}) [/itex] as [itex] \delta(\sqrt{x}-0) [/itex]
 
  • #9
Aside: I know the hand-wavy argument is clear, but does anyone know of a source that actually defines the result of composing a distribution with a function of some sort?
 
  • #10
so assuming i made the substitution right and that was what i was supposed to do, i still don't see how i proved that it equals zero.

is that supposed to say if i plug in zero for x i get zero?

and even if i treat [itex]
\delta(\sqrt{x})
[/itex] as [itex]
\delta(\sqrt{x}-0)
[/itex]
i still don't have a function to plug zero into.

i am still quite confused about what i am trying to do.
 
  • #11
Hurkyl said:
Aside: I know the hand-wavy argument is clear, but does anyone know of a source that actually defines the result of composing a distribution with a function of some sort?

Section 7.4.d Composition of [itex]\delta[/itex] with a function, from Mathematics for Physics and Physicists by Walter Appel.

The idea, as usual, is to to use the distribution obtained by integrating a locally integrable function against test functions to motivate a more general definition.

1. Use locally integrable [itex]g[/itex] and integration to generate a distribution [itex]G[/itex].

2. Use locally integrable [itex]g \circ f[/itex] and integration to generate a distribution denoted by [itex]G \circ f[/itex].

3. If [itex]x[/itex] is the integration variable in 2., make the substitution [itex]y = f(x)[/itex].

4. Relate [itex]G \circ f[/itex] to [itex]G[/itex].

5. Use 4. to motivate the definition of [itex]T \circ f[/itex] in tems of an arbitrary distribution [itex]T[/itex] and differentiable and bijective function [itex]f[/itex].

If I get time tomorrow, I might type in the details.
 
Last edited:

Related to Starting with the definition of the Dirac delta function,

What is the definition of the Dirac delta function?

The Dirac delta function, denoted by δ(x), is a mathematical function that is defined as zero everywhere except at x = 0, where it is infinite in magnitude but has an area of 1. It is often referred to as an "impulse" function due to its sharp peak at x = 0.

What is the physical interpretation of the Dirac delta function?

The Dirac delta function is often used to model point-like sources or point masses in physics. It represents a perfectly localized concentration of energy or mass at a single point in space.

How is the Dirac delta function used in mathematical equations?

The Dirac delta function is used as a distribution in mathematical equations, meaning it is not a traditional function that can be evaluated at every point, but rather a generalized function that is defined by its properties. It is commonly used in integration, Fourier transforms, and differential equations.

What are the properties of the Dirac delta function?

The Dirac delta function has several important properties, including: 1) δ(x) = 0 for all x ≠ 0, 2) ∫δ(x)dx = 1, 3) δ(ax) = δ(x)/|a| where a is a constant, and 4) δ(x) = δ(-x), meaning it is an even function.

What are some real-world applications of the Dirac delta function?

The Dirac delta function has many applications in physics, engineering, and mathematics. Some examples include modeling the electric potential of a point charge, analyzing impulse responses in signal processing, and solving partial differential equations in fluid dynamics.

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