Generalized functions (distributions) problem - Mathematical physics

[Sigurdsson]

Homework Statement

Find a distribution $$g_n$$ which satisfies
$$g'_n(x) = \delta(x - n) - \delta(x + n)$$
and use it to prove
$$\lim_{n \to \infty} \frac{\sin{nx}}{\pi x} = \delta(x)$$

Homework Equations

Nothing relevant comes up at the moment.

The Attempt at a Solution

Well the first part is pretty easy I think. The distribution would be
$$g_n(x) = \theta(x - n) - \theta(x + n) = \left\{ \begin{array}{l l} -1 & \quad |x| < n \\ 0 & \quad |x| \geq n \\ \end{array} \right.$$


The limit will indeed resemble a delta function when $$n$$ goes to infinity and π is probably just a normalization constant. But applying the two Heaviside functions to solve this has got me stumped.

P.S. Gotta catch some sleep, I will be back in 7 hours hopefully with some ideas to solve this.

Cheers

 

[vela]

g' looks like the Fourier transform of sine.

 

[Sigurdsson]

Thanks Vela! That did the trick, I'll post the full solution later. :)

 

[Sigurdsson]

Right guys so here it is, me and my buddies came up with this solution to the problem.

So we have established that our distribution should be
$$g_n(x) = \theta(x - n) - \theta(x + n)$$
Like Vela said, the Fourier transform of this sine resembles our distribution differentiated. So that means we have (using Euler equation for sine)
$$\frac{d}{d\xi} \mathcal{F} \left\{ \frac{\sin{(nx)}}{\pi x} \right\}(\xi) = \frac{d}{d \xi} \int \frac{1}{\pi x} e^{i \xi x} \left( \frac{e^{inx} - e^{-inx} }{2i} \right) dx = \int \frac{i x}{2 i \pi x} e^{i\xi x} (e^{inx} - e^{-inx}) dx$$
$$= \frac{1}{2\pi} \int e^{ix(\xi +n)} - e^{ix(\xi - n)} dx = \frac{2\pi}{2\pi} \left( \delta(\xi + n) - \delta(\xi - n) \right) = \frac{d}{d\xi} \left( \theta(\xi + n) - \theta(\xi - n) \right) = \frac{d}{d\xi} (- g_n(\xi))$$
If we look at the limit of our distribution. You can easily see that it will become unity. (You can plot those heaviside functions and see for yourself).
$$\lim_{n \to \infty} (- g_n(\xi)) = 1$$
Using inverse Fourier-Transform we have
$$\mathcal{F}^{-1} \mathcal{F} \left\{ \frac{\sin{(nx)}}{\pi x} \right\}(\xi) = \mathcal{F}^{-1} \{1\}(x) = \delta(x)$$
Which ultimately proves what was supposed to prove
$$\lim_{n \to \infty} \frac{\sin{(n x)}}{\pi x} = \delta(x)$$


gg

 

[paranormal]

!I would like to offer some guidance on how to approach this problem. First, let's define the Dirac delta function as a distribution, denoted by δ(x), which satisfies the following properties:

1. δ(x) = 0 for x≠0
2. ∫ δ(x)dx = 1
3. For any smooth function ƒ(x), ∫ ƒ(x)δ(x)dx = ƒ(0)

Next, let's use the fact that the derivative of the Heaviside step function, θ(x), is the Dirac delta function, δ(x), and that the derivative of a constant function is 0. Therefore, we can rewrite the given equation as:

g'_n(x) = -θ'(x - n) + θ'(x + n)

Now, let's use the properties of the delta function to solve for g_n(x):

g_n(x) = -θ(x - n) + θ(x + n) + C

where C is a constant. To determine the value of C, we can use the fact that g_n(x) must be continuous at x = n and x = -n. This means that g_n(n) = g_n(-n). Substituting in the above equation, we get:

-θ(n - n) + θ(n + n) + C = -1
-θ(-n - n) + θ(-n + n) + C = 0

Simplifying, we get:

-θ(n) + θ(2n) + C = -1
θ(n) + C = 0

Solving for C, we get C = 1. Therefore, the final distribution is:

g_n(x) = -θ(x - n) + θ(x + n) + 1

Now, to prove the given limit, we can use the fact that the derivative of sin(nx)/πx is equal to δ(x - n) - δ(x + n), which is the same as g_n(x). Therefore, we can write:

∫ g_n(x) sin(nx)/πx dx = ∫ (δ(x - n) - δ(x + n)) sin(nx)/πx dx

Using the properties of the delta function, we can rewrite this as:

∫ g_n(x) sin(nx)/πx dx