Prove that derivative of the theta function is the dirac delta function

In summary: This is a rigorous way to do the problem, but it's not necessary to do so. It's also not necessary to separate the integral into positive and negative x cases, as long as you understand what is happening.
  • #1
demonelite123
219
0
let θ(x-x') be the function such that θ = 1 when x-x' > 0 and θ = 0 when x-x' < 0. Show that d/dx θ(x-x') = δ(x - x').

it is easy to show that d/dx θ(x-x') is 0 everywhere except at x = x'. To show that d/dx θ(x-x') is the dirac delta function i also need to show that the integral over the real line of d/dx θ(x-x') = 1.

this is what i tried: ∫ d/dx θ(x-x') dx' = d/dx ∫ θ(x-x') dx' = d/dx ∫ 1 dx' but now i am a little stuck and not sure if this is the way to go or not. my plan was to show that the integral equals x so that d/dx (x) = 1 like we wanted. but it seems like the bottom limit of integration will be x and i'll end up getting -1 instead of 1. can anyone give me some advice on this problem? thanks.
 
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  • #2
This is one of those problems that tends to be really easy if you think of them in a rigorous fashion rather than a heuristic "intuitive" fashion.

What foundations are you using for the delta function and derivatives and such? When working with tempered distributions, the derivative of a distribution F is the (unique) distribution which satisfies, for all test functions [itex]\varphi[/itex],
[tex]\int_{-\infty}^{+\infty} F'(x) \varphi(x) \, dx = -\int_{-\infty}^{+\infty} F(x) \varphi'(x) \, dx[/tex]​
 
  • #3
Just think about what is [itex]f(x) = \int_{-\infty}^{x} \delta(t) dt [/itex] for each [itex]x \ne 0[/itex]. You should obvious separate it into positive and negative [itex]x[/itex] case. Then the fundamental theorem of calculus applies.
 
  • #4
mathfeel said:
Just think about what is [itex]f(x) = \int_{-\infty}^{x} \delta(t) dt [/itex] for each [itex]x \ne 0[/itex]. You should obvious separate it into positive and negative [itex]x[/itex] case. Then the fundamental theorem of calculus applies.

ah that's a similar idea to what i did. thanks for your answer! so would this be a more rigorous way to do this problem? i know that taking the derivative operator out of the integral isn't being very careful but the physics book seemed to do that all the time so i thought it would be "ok" since i wan't doing mathematics here.
 
  • #5


First, let's clarify the definition of the dirac delta function. The dirac delta function, denoted as δ(x), is a mathematical concept that represents an infinitely tall and narrow spike at x = 0, with an area under the curve of 1. It is also commonly defined as a function that is 0 everywhere except at x = 0, where it is infinite. Its integral over the real line is equal to 1.

Now, let's look at the given function, θ(x-x'). As stated, it is equal to 1 when x-x' > 0 and 0 when x-x' < 0. This means that it is essentially a step function, with a discontinuity at x = x'. We can rewrite θ(x-x') as H(x-x'), where H(x) is the Heaviside step function. This function is defined as 0 for x < 0 and 1 for x > 0. Notice that when x = x', H(x-x') is equal to 1.

Now, let's take the derivative of θ(x-x'). Using the chain rule, we get d/dx θ(x-x') = d/dx H(x-x'). Since H(x-x') is a step function, its derivative is equal to the dirac delta function, δ(x-x'). Therefore, we have shown that d/dx θ(x-x') = δ(x-x').

To show that the integral of d/dx θ(x-x') over the real line is equal to 1, we can use the definition of the dirac delta function. Recall that the dirac delta function has the property that ∫ δ(x) dx = 1. Using this property, we can rewrite the integral as ∫ d/dx θ(x-x') dx = ∫ δ(x-x') dx. Since we know that d/dx θ(x-x') = δ(x-x'), we can substitute and get ∫ δ(x-x') dx = ∫ δ(x-x') dx. This is true by the definition of the dirac delta function, so we have shown that the integral of d/dx θ(x-x') is equal to 1.

In conclusion, we have proven that the derivative of the theta function is the dirac delta function, by showing that d/dx θ(x-x') = δ(x-x') and that the integral of d
 

1. What is the theta function?

The theta function is a mathematical function used in number theory and Fourier analysis. It is denoted by the Greek letter theta (θ) and is defined as θ(x) = ∑n=-∞ eπixn2.

2. What is the derivative of the theta function?

The derivative of the theta function is given by θ'(x) = ∑n=-∞ 2πinx eπixn2.

3. What is the dirac delta function?

The dirac delta function is a mathematical function that is defined as zero for all values except at x=0, where it tends to infinity. It is often used in physics and engineering to model point-like interactions or impulses.

4. How is the derivative of the theta function related to the dirac delta function?

The derivative of the theta function is directly related to the dirac delta function. Specifically, θ'(x) = ∑n=-∞ 2πinx eπixn2 = δ(x), where δ(x) represents the dirac delta function.

5. Why is it important to prove that the derivative of the theta function is the dirac delta function?

This proof is important because it establishes a fundamental relationship between two important mathematical functions. It also has various applications in physics, engineering, and signal processing, making it a crucial concept for understanding and solving real-world problems.

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