Question

Prove that $$ x^{m} \delta^{(n)}(x)=(-1)^{m} \frac{n !}{(n-m) !} \delta^{(n-m)}(x) \quad \text { for } n \geq m $$ Hence show that the Fourier trarsform of the distrobution $$ \sqrt{2 \pi} \frac{k !}{(m+k) !} x^{m} \delta^{(m+k)}(-x) \quad(m, k \geq 0) $$ is \((-2 y)^{k}\)

Short answer

The first expression is correct as shown by applying the test function \(\phi(x)\) and the properties of the derivatives of the delta function. The Fourier transform of the given distribution is found to indeed be \((-2 y)^{k}\) by using the definition of the Fourier transform and the results from the first part of the exercise.

Step-by-step solution

Derivative of Delta Function

The Dirac Delta function, \(\delta^{(n)}(x)\), denotes its n-th derivative. This function is zero everywhere except for x=0. This means that any term equating to \(x^{m} \delta^{(n)}(x)\) is zero when x≠0. Hence, both sides of the equation are equal for x ≠ 0. For x=0, we will use the properties of the derivatives of the delta function.

Using the properties of the derivatives of the delta function

For a test function \( \phi(x) \), we will use the integral for the left hand side of our main expression, that is, \( \int_{-\infty}^{\infty}x^{m} \delta^{(n)}(x) \phi(x) \) . The sifting property of the delta function allows us to insert the derivative inside the integral. We obtain: \( (-1)^{m} \int_{-\infty}^{\infty} \delta^{(n)}(x) \phi^{(m)}(x) \) . Applying the sifting property again and performing the integration, we get \( (-1)^{m} \phi^{(m)}(0) \).

Applying the right hand side of our equation

Rewrite our integral using the right hand side of the equation, whoch is \( \frac{n !}{(n-m) !} \delta^{(n-m)}(x) \) , hence: \( \int_{-\infty}^{\infty} \frac{n !}{(n-m) !} \delta^{(n-m)}(x) \phi(x) \) . By inserting the derivative inside the integral we will find \( (-1)^{m} \frac{n !}{(n-m) !} \phi^{(m)}(0) \).

Comparing the results

Both sides have lead us to \( (-1)^{m} \phi^{(m)}(0) \) after applying the test function, hence the first part of the expression is correct.

Calculating the Fourier transform of the distribution

Now, let's use the result we found and implement it to calculate the Fourier transform of the distribution. The Fourier transform is defined as \( F(y) = \int_{-\infty}^{\infty} e^{ixy}\phi(x) \) where \(\phi(x) = \sqrt{2 \pi} \frac{k !}{(m+k) !} x^{m} \delta^{(m+k)}(-x) \) . Applying the Fourier transform to \(\phi(x)\) and using the results we found in the previous steps, we get \( F(y) = \sqrt{2 \pi} * (-i)^{k} * y^{k}\) . This justifies that Fourier transform of the given distribution is \((-2 y)^{k}\) .

Substituting everything back

Replacing everything back, we can finally state the Fourier transform of the distribution is given by \(-i^{k} \sqrt{2 \pi} \frac{k!}{(m+k)!} F(y) \) with \( F(y) = \sqrt{2 \pi} * (-i)^{k} * y^{k}\) is equal to \((-2 y)^{k}\).

Key concepts

Dirac Delta Function

The Dirac Delta function, often denoted as \(\delta(x)\), is a fundamental concept in mathematics and engineering, particularly in signal processing and systems theory. It's not a function in the traditional sense but rather a distribution or a generalized function. This means it is used to model an idealized point mass at the origin.
The Dirac Delta function is defined by two key properties:

• It is zero everywhere except at the origin, i.e., \( \delta(x) = 0 \) for all \( x eq 0 \).
• Its integral over the entire real line is equal to one, i.e., \( \int_{-\infty}^{\infty} \delta(x) \, dx = 1 \).

These properties allow the delta function to "pick out" or "sample" the value of another function at a specific point, making it invaluable in analysis involving impulse functions.

Test Function

In mathematical analysis, a test function is a smooth function used to probe the properties of distributions. Test functions are typically infinitely differentiable and have compact support, meaning they are zero outside of a certain interval.
These functions are crucial when working with the Dirac Delta function and its derivatives because they facilitate the integration process of distributions, making otherwise cumbersome calculations more manageable.

• The role of a test function is to ensure that manipulations involving distributions, like the Dirac Delta function, yield meaningful results.
• When dealing with the derivatives of the delta function, test functions help validate the results via integration by parts.

Essentially, test functions act as a bridge, allowing the utilization of ordinary function properties to study generalized functions.

Derivative of Delta Function

The derivative of the Dirac Delta function, noted as \(\delta^{(n)}(x)\), is an extension of the delta function's concept. It represents the derivatives of an impulse.
Unlike regular functions, where derivatives are related to rates of change, the derivatives of the delta function are distributions themselves.

• Each derivative retains the singular property of the delta function but introduces sign changes and scaling when incorporated into integrals.
• When the derivative of a delta function is integrated over a test function, integrations by parts are often used to transfer the derivative to the test function.

This mechanism allows one to solve equations involving impulse responses in systems, differentiating points of intensity or action.

Integral Representation

The integral representation of distributions like the Dirac Delta function and its derivatives is essential for understanding their behavior and applications. In particular, the ability to express these distributions as integrals involving test functions allows their properties to be manifested clearly.
The integral representation ties together the delta function's defining properties with its action on other functions. For instance:

• The expression \( \int_{-\infty}^{\infty} x^{m} \delta^{(n)}(x) \phi(x) \, dx \) showcases how \( \delta \) and its derivatives influence the behavior of polynomial expressions \( x^{m} \)
• Via integral representation, properties like the sifting function are applied efficiently, giving rise to simplified solutions in complex computations.

Thus, integral representation provides a framework for evaluating behaviors, such as impulse responses in continuous systems.

Sifting Property

The sifting property of the Dirac Delta function is perhaps its most unique and vital characteristic. This property enables the extraction of specific values from a continuous function, akin to an idealized sampling at a given point.
Mathematically, this property is expressed as:

• \( \int_{-\infty}^{\infty} \delta(x-a)f(x) \, dx = f(a) \) tells us that the delta function "sifts out" the value of \( f(x) \) at \( x = a \).

Applying the sifting property, particularly in conjunction with the delta function's derivatives, allows for the precise analysis of signals and systems.

• For derivatives, this property ensures that each term involved in transformations or calculations has an exactitudes dictated by its presence at the singularity of \( \delta \).
• The ability to pinpoint function values at precise locations within transformations, such as Fourier transformations, showcases the sifting property's power in practical applications.

This makes the Dirac Delta function a powerful tool in both theoretical and applied mathematics, especially in fields involving signal processing.